Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Engineering LibreTexts

11: Rigid Body Kinematics

  • Last updated
  • Save as PDF
  • Page ID 50609

  • Jacob Moore & Contributors
  • Pennsylvania State University Mont Alto via Mechanics Map
  • 11.1: Fixed-Axis Rotation in Rigid Bodies Introduction to rotational kinematics: angular position, velocity and acceleration equations; determining angular velocity and acceleration of a point on a body rotating about a fixed axis. Includes worked examples.
  • 11.2: Belt- and Gear-Driven Systems Rotational kinematics of belt-driven pulley systems, simple gear systems, and compound gear trains. Includes worked examples.
  • 11.3: Absolute Motion Analysis Absolute motion analysis of extended bodies undergoing general planar motion. Includes worked examples.
  • 11.4: Relative Motion Analysis Relative motion analysis of extended bodies undergoing general planar motion. Includes worked examples.
  • 11.5: Rotating Frame Analysis Discussion of reference frames that translate and/or rotate with the motion of an object; derivation of kinematics equations for rotating frame analysis (analysis of multiple moving objects that are not pinned together). Includes worked examples. **Video lecture not yet available.**
  • 11.6: Chapter 11 Homework Problems

Academia.edu no longer supports Internet Explorer.

To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to  upgrade your browser .

Enter the email address you signed up with and we'll email you a reset link.

  • We're Hiring!
  • Help Center

paper cover thumbnail

Dynamics of Rigid Bodies (Problems and Solutions)

Profile image of ROLAND THIRDTHE PACHECO

2023, Roland Thirdthe V. Pacheco

In the realm of engineering mechanics, where the synergy of mathematics and physics forms the bedrock of understanding, emerges a beacon of assistance for aspiring students. This excerpt, a glimpse into the forthcoming volume titled "Dynamics of Rigid Bodies, Problems and Solutions," is a testament to unwavering optimism. Crafted with the conviction that knowledge shared is a torch passed, this work aims to illuminate the intricate pathways of learning for students in the realm of engineering. As these pages unfold, so too do the solutions to problems that have often proven formidable. It is with sincere hope and unyielding trust that this contribution shall empower students, offering them a guiding light through the complexities of their academic journey.

RELATED PAPERS

La Nuova Bussola Quotidiana

Aurelio Porfiri

Materials Today: Proceedings

Physical Review D

Gastao Krein

Diagnostics

Teodora Bujaroska

Kansas Agricultural Experiment Station Research Reports

Randall Currie

Olle Kejonen

L'Endocrinologo

Marco Centanni

International Journal of Innovative Research in Engineering & Management (IJIREM)

IJIREM JOURNAL

The Annals of thoracic surgery

Rémi HERVOCHON

BioTechniques

alberto benguria

Chemistry – A European Journal

Viatcheslav Jouikov

Cabezudo, Baltasar

Genes & Development

Michael Downes

International Journal of Stomatology & Occlusion Medicine

ashima behl

Lingua Rima: Jurnal Pendidikan Bahasa dan Sastra Indonesia

Fahriza Ramadhan

Kristanti Parisihni, drg, M.Kes

Modern Biotechnologies – Solutions to the Challenges of the Contemporary World. Symposium Proceedings

Ludmila B A L A N Batir

Basic and Applied Education Research Journal

Journal of Andrology

Calogero Stelletta

Hasyim Asy'ari

relatos sobre cómo investigar puede des-velar-nos

Griselda Amuchastegui

International Journal of Engineering and Computer Science

Nasurudeen Ahamed N

RELATED TOPICS

  •   We're Hiring!
  •   Help Center
  • Find new research papers in:
  • Health Sciences
  • Earth Sciences
  • Cognitive Science
  • Mathematics
  • Computer Science
  • Academia ©2024

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Physics LibreTexts

Sample Problems

  • Last updated
  • Save as PDF
  • Page ID 63408

  • Tom Weideman
  • University of California, Davis

All of the problems below have had their basic features discussed in an "Analyze This" box in this chapter. This means that the solutions provided here are incomplete, as they will refer back to the analysis performed for information (i.e. the full solution is essentially split between the analysis earlier and details here). If you have not yet spent time working on (not simply reading!) the analysis of these situations, these sample problems will be of little benefit to your studies.

Problem 5.1

A bug stands on the outer edge of a turntable as it begins to spin, accelerating rotationally in the horizontal plane from rest at a constant rate. The bug is held on the turntable by static friction, but as the turntable spins ever faster, this will not remain the case forever.  The turntable, which has a radius of \(0.40m\), has its rotational speed increase at a steady rate from rest, and reaches a speed of \(\omega_1=1.2\frac{rad}{s}\) after its first full revolution.

  • Find the linear speed of the bug after the turntable makes \(n\) full revolutions.
  • Find the coefficient of static friction if the bug falls off after the turntable makes \(n\) full revolutions.

a. In the analysis , we indicated a kinematics relationship between \(\alpha\), \(\theta\), and \(\omega\). The angular acceleration is constant, so it does not depend on the number of revolutions, and for \(n\) revolutions, \(\theta=2\pi n\), so the angular velocity after \(n\) revolutions is:

\[\omega_n^2=2\alpha\left(2\pi n\right)=\omega_1^2 n\;\;\;\Rightarrow\;\;\;\omega_n=\omega_1\sqrt{n}\nonumber\]

The linear speed is the angular speed multiplied by the radius, so:

\[v_n=r\omega_n=r\omega_1\sqrt{n}=\left(0.40m\right)\left(1.2\frac{rad}{s}\right)\sqrt{n}=\left(0.48\frac{m}{s}\right)\sqrt{n}\nonumber\]

b. In the analysis we derived a formula for the coefficient of static friction in terms of the angular velocity and acceleration. Incorporating what we found above in terms of the number of full revolutions, we get:

\[\mu_s=\frac{r\sqrt{\omega_n^4+\alpha^2}}{g}=\frac{r\sqrt{n^2\omega_1^4+\left(\frac{\omega_1^2}{4\pi }\right)^2}}{g}=\frac{r\omega_1^2}{g}\sqrt{n^2+\frac{1}{16\pi^2}}=0.059\sqrt{n^2+0.0063}\nonumber\]

Notice that unless the angular acceleration is very large so that the bug falls off very quickly – in much less than a full revolution (\(n\ll 1\)) – virtually all of the static friction force goes into maintaining the centripetal acceleration, and only a very tiny fraction of the total static friction is involved with speeding up the bug.

Problem 5.2

The Blort Corporation makes a special widget that consists of a uniform disk pivoted around an axle at the end of a rod of negligible mass, which in turn rotates about its other end. This widget has two settings: It can be set in the "locked" position so that the disk does not rotate around its axle, or the "free" position so that the disk rotates frictionlessly about the axle. The difference these settings have on the motion of the disk as the rod rotates is depicted in the figure below.

locked_or_free_axis.png

An engineer for a company that uses the Blort widgets in their manufacturing wants to make sure that the power output of the motor that turns the rod automatically adjusts so that the rod's rotation is the same whether the axle is in the fixed or free setting. The specifications of the widget indicates that the rod's length is equal to the diameter of the disk. By what factor must the power output of the motor be adjusted between these two settings?

In the analysis , we calculated the kinetic energies for a given rotational speed for each of these settings. The power from the motor goes directly into the kinetic energy of the widget, so the ratio of the kinetic energies will match the ratio of the power outputs:

\[\frac{P_{locked}}{P_{free}}=\frac{ {KE}_{locked}}{ {KE}_{free}} = \dfrac{\frac{1}{4}M\left(R^2+2L^2\right)\omega^2}{\frac{1}{2}ML^2\omega^2} = 1+\frac{1}{2}\left(\frac{R}{L}\right)^2\nonumber\]

The length of the rod is the diameter of the disk, so it is twice the radius, giving:

\[\frac{P_{locked}}{P_{free}} = 1.125\nonumber\]

Problem 5.3

A solid uniform sphere starts from rest and rolls down a flat ramp without slipping.

ball_rolls_down_ramp.png

The sphere descends a vertical distance of \(3.6m\) by the time it reaches the bottom, and it takes \(6.6s\) to make the journey. Find the angle that the ramp makes with the horizontal.

We know the final speed of the sphere from our analysis :

\[v_f=\sqrt{\frac{10}{7}gh}\nonumber\]

The sphere started from rest, and as noted in the analysis, the acceleration is constant, so the average velocity is:

\[v_{ave}=\frac{v_o+v_f}{2}=\sqrt{\frac{5}{14}gh}\nonumber\]

This number, along with the time, gives us the straight-line (along the ramp) distance traveled by the ball:

\[d=v_{ave}t=\sqrt{\frac{5}{14}gh}\;t\nonumber\]

We now have the hypothenuse and opposite side of a right triangle, so we can get the angle:

\[\theta = \sin^{-1}\left(\frac{h}{d}\right)=\sin^{-1}\left(\sqrt{\frac{14h}{5gt^2}}\right)=\sin^{-1}\left(\sqrt{\frac{14\left(3.6m\right)}{5\left(9.8\frac{m}{s^2}\right)\left(6.6s\right)^2}}\right)=8.8^o\nonumber\]

Problem 5.4

A solid and a hollow sphere roll without slipping simultaneously (one behind the other) down a ramp and around a loop-de-loop. The radii of the spheres are negligible compared to the radius of the loop.

two_spheres_around_loop.png

Both spheres are released simultaneously from rest, and both barely make it around the loop. Find which sphere is in front of the other, and the ratio of their starting heights.

In the analysis we found that if they have the same velocity, the two spheres will have different kinetic energies. We also found the speed either sphere must have in order to get all the way around the loop. Plugging this value into the kinetic energies of the spheres tells us how much kinetic energy they must have to make it around:

\[\left.\begin{array}{l} KE_{solid}=\frac{7}{10}mv^2 \\ v=\sqrt{gR} \end{array}\right\}\;\;\; KE_{solid}=\frac{7}{10}mgR\nonumber\]

\[\left.\begin{array}{l} KE_{hollow}=\frac{5}{6}mv^2 \\ v=\sqrt{gR} \end{array}\right\}\;\;\; KE_{solid}=\frac{5}{6}mgR\nonumber\]

Referencing zero gravitational potential energy at the bottom of the loop, at the top of the loop, the spheres alsoi have a potential energy of \(2mgR\). Given that they start from rest, they start with only potential energy, so that equals their total energy:

\[E_{tot\;hollow}=mgh_{hollow}=mg\left(2R+\frac{5}{6}R\right)=\frac{17}{6}mgR\nonumber\]

\[E_{tot\;solid}=mgh_{solid}=mg\left(2R+\frac{7}{10}R\right)=\frac{27}{10}mgR\nonumber\]

The hollow sphere needs to start higher, and since they start simultaneously from rest, the leading sphere is the solid one. The ratio of their starting heights is:

\[\frac{h_{hollow}}{h_{solid}} = \frac{85}{81}\nonumber\]

Problem 5.5

One end of a massless rope is wound around a uniform solid cylinder, while the other end passes over a massless, frictionless pulley and is attached to a hanging block, as in the diagram below. The block is released from rest, pulling the cylinder along the horizontal surface such that it rolls without slipping.

block_pulls_spool_that_rolls.png

The cylinder and block are both weigh \(22N\). Find the tension in the string.

In the analysis , we found the final velocity of the cylinder after the block drops a distance \(h\). Plugging in equal masses for the block and cylinder gives, and solving for the velocity of the block using, \(v_b=2v_c\):

\[v_f=2\sqrt{\frac{4m}{8m+3m}gh}=\sqrt{\frac{16}{11}gh}\nonumber\]

The acceleration of the block is constant, so from the kinematics equation with no time variable, we can get the acceleration:

\[v_f^2-v_o^2=2ay\;\;\;\Rightarrow\;\;\; a=\frac{v_f^2}{2h}=\frac{8}{11}g\nonumber\]

The acceleration is a result of the net force, which is the vector sum of the downward gravitational force and the upward tension force, so setting upward as the positive direction (making the acceleration negative), we get:

\[T-mg=ma\;\;\;\Rightarrow\;\;\;T=m\left(g+a\right)=m\left(g-\frac{8}{11}g\right)=\frac{3}{11}mg=\frac{3}{11}\left(22N\right)=6N\nonumber\]

Problem 5.6

Two disks are cut out of the same material, as shown in the diagram below. They are pivoted around stationary axles such that the two disks lie in the vertical plane, with their outer rims pinching a massless rope between them. The rope is pulled downward, causing both disks to turn without the rope slipping.

two_disks.png

The smaller disk has one-third the radius of the larger disk. As the rope is pulled, power is delivered to the two-disk system. Find the fraction of the total power delivered to the larger disk.

To find the fraction of the power delivered, we only need to figure out the ratio of the energy of the small disk to the large disk at a given speed.  This ratio is:

\[\frac{ {KE}_{small}}{ {KE}_{large}}=\dfrac{\frac{1}{2}I_{small}\omega_{small}^2}{\frac{1}{2}I_{large}\omega_{large}^2}=\left(\frac{I_{small}}{I_{large}}\right)\left(\frac{\omega_{small}}{\omega_{large}}\right)^2\nonumber\]

We found these ratios in terms of the ratios of the radii of the disks in the analysis .

\[\left.\begin{array}{l} \frac{I_{small}}{I_{large}}=\left(\frac{r}{R}\right)^4 \\ \frac{\omega_{small}}{\omega_{large}}=\frac{R}{r} \end{array}\right\} \;\;\; \frac{ {KE}_{small}}{ {KE}_{large}}= \left(\frac{r}{R}\right)^2=\left(\frac{1}{3}\right)^2=\frac{1}{9}\nonumber\]

Nine times as much power goes to the large disk as the small disk, which means that 90% of the total power delivered by the pulled rope is going to the large disk.

Problem 5.7

One end of a uniform metal thin rod is welded to the outer edge of a metal disk. The masses of these two objects are the same, and the length of the rod is equal to the diameter of the disk. The disk is suspended on a frictionless axle positioned at its center, and the rod is released from rest from a horizontal orientation and allowed to swing down to the vertical position.

swinging_disk_and_rod.png

The linear speed of the open end of the rod at the bottom of the swing is measured to be \(3.0\frac{m}{s}\). The pendulum is then removed from the axle and is swung in the same manner (from rest horizontally) with the open end of the rod now attached to the axle (so the disk is swinging down). Find the linear speed of the bottom edge of the disk when it gets to the bottom of the swing.

Let's start by using our result from the analysis to determine what is given. We know that the linear speed of the bottom of the rod is the angular velocity multiplied by the distance to the axle, so:

\[v_{rod\;end}=r\omega\;\;\;\Rightarrow\;\;\;3.0\frac{m}{s}=3R\omega_{rod\;swings}\nonumber\]

We can solve for the value of \(R\) here, but as we will soon see, this is not necessary. We do, however, have to follow all the same steps from the analysis for the new setup.  First, when pivoted at the open end of the rod, the center of mass of the whole object descends a distance of \(2R\) (twice as far as the previous case), giving:

\[\Delta U = -2mg\left(2R\right)=-4mgR\nonumber\]

The moment of inertia also changes. This time we have the rod about its end and the disk extended by the parallel axis theorem:

\[I_{tot} = I_{rod}+I_{disk} = \frac{1}{3}m\left(2R\right)^2+\left[\frac{1}{2}mR^2+m\left(3R\right)^2\right] = \frac{65}{6}mR^2\nonumber\]

Putting this into the energy conservation gives:

\[{KE}_f=-\Delta U \;\;\;\Rightarrow\;\;\; \frac{1}{2}\left(\frac{65}{6}mR^2\right)\omega^2=4mgR\;\;\;\Rightarrow\;\;\; \omega=\sqrt{\frac{48g}{65R}}\nonumber\]

This is a bit slower than the previous case. The ratio of the two angular velocities are:

\[\frac{\omega_{disk\;swings}}{\omega_{rod\;swings}}=\sqrt{\frac{58}{65}}\nonumber\]

But we are interested in the linear velocity of the bottom of the disk. This is a distance of \(4R\) from the axle, so using the result above, its linear velocity is:

\[v_{disk\;bottom}=4R\omega_{disk\;swings}=4R\left(\sqrt{\frac{58}{65}}\omega_{rod\;swings}\right)=\frac{4}{3}\sqrt{\frac{58}{65}}\left(3R\omega_{rod\;swings}\right)=\sqrt{\frac{928}{585}}\left(3.0\frac{m}{s}\right)=3.8\frac{m}{s}\nonumber\]

Problem 5.8

A small marble is attached to the end of a thin rigid rod with an equal mass, whose other end is held fixed at the origin. The rod starts at rest in the \(x-y\) plane, and makes an angle \(\theta\) up from the \(x\)-axis, as shown in the diagram.  There is no gravity present, but the marble (not the rod) is subjected to a force from a potential energy field given by:

\[ U\left(x,y\right)=\beta x y+U_o \;,\;\;\;\; \beta=constant>0\nonumber \]

marble_on_rod.png

The values of the variables given above are:

\[\theta=30^o\;,\;\;\;\; \text{mass of marble}=0.45kg\;,\;\;\;\; \beta=1.2\frac{J}{m^2}\nonumber\]

  • Find the magnitude and direction of the angular acceleration when the rod is released. Express the direction of this acceleration both as a unit vector and as either clockwise or counterclockwise from the perspective of this diagram.
  • Find the maximum angular velocity attained by the rod, and the orientation angle of the rod when this maximum is reached.

a. As usual, most of the heavy-lifting in this problem was already done in the analysis . Using the expression for the angular acceleration derived in the analysis, we have:

\[\vec\alpha=-\frac{3\beta\cos2\theta}{4m}\;\hat k=-\frac{3\left(1.2\frac{J}{m^2}\right)\cos60^o}{4(0.45kg)}\;\hat k=-1.0\frac{rad}{s^2}\hat k\nonumber\]

This vector is in the \(-\hat k\) direction, which points into the page, as we are using a right-handed coordinate system. From the RHR, this direction is clockwise from the perspective looking at the diagram.

b. Note that the linear velocity is a maximum when the rotational velocity is a maximum, so if we write the rotational velocity as a function of \(\theta\), we just need to do calculus to find where the maximum occurs:

\[0=\frac{d\omega}{d\theta}=\frac{d\omega}{dt}\frac{dt}{d\theta}=\frac{\alpha}{\omega}\;\;\;\Rightarrow\;\;\;\alpha=0\nonumber\]

From the result for the angular acceleration, we see that the extrema occur at \(\cos2\theta=0\), so \(\theta = \pm 45^o\) or \(\theta = \pm 135^o\). All of these angles satisfy \(\left|x\right| = \left|y\right|.\) Next we need to determine which ones correspond to the maximum speed. Clearly the maximum kinetic energy occurs when the potential energy is a minimum, and looking at the potential energy function, this occurs when either \(x\) or \(y\) (but not both!) is negative. These two cases correspond to \(\theta = -45^o\) and \(\theta = 135^o\).  The other two angles correspond to maximum potential energies, but since the marble starts with zero kinetic energy at a lower potential energy, it can never reach these points. Therefore the only place where the marble (which starts at \(\theta=30^o\) and accelerates clockwise) can reach a maximum speed is \(\theta=-45^o\).

Let's call the length of the rod \(L\), as in the analysis.  When the marble gets to \(\theta=-45^o\), the potential energy is:

\[x=-y=\frac{1}{\sqrt 2}L;\;\;\Rightarrow\;\;\;U=-\frac{1}{2}\beta L^2+U_o\nonumber\]

Initially we had:

\[\left.\begin{array}{l}x=L\cos30^o=\frac{\sqrt 3}{2}L \\ y=L\sin30^o=\frac{1}{2}L \end{array}\right\} \;\;\;\;U=\frac{\sqrt 3}{4}\beta L^2+U_o\nonumber\]

The object starts from rest, so its final kinetic energy is just equal to the amount of potential energy lost:

\[KE=-\Delta U\;\;\;\Rightarrow\;\;\;\frac{1}{2}I_{tot}\omega^2=\left(\frac{\sqrt 3}{4}\beta L^2+U_o\right)-\left(-\frac{1}{2}\beta L^2+U_o\right)=\left(\frac{\sqrt 3+2}{4}\right)\beta L^2\nonumber\]

Plugging in for the moment of inertia of the whole object as found in the analysis, we get:

\[\frac{1}{2}\left(\frac{4}{3}mL^2\right)\omega^2=\left(\frac{\sqrt 3+2}{4}\right)\beta L^2\;\;\;\Rightarrow\;\;\;\omega = \sqrt{\frac{3}{8}\left(\sqrt 3+2\right)\left(\frac{1.2\frac{J}{m^2}}{0.45kg}\right)}=1.9\frac{rad}{s}\nonumber\]

Problem 5.9

Returning to the physical system of problem 5.6, we consider a new question. As before, the cylinder rolls without slipping, and the masses are equal (though here their exact values are not known).

block_pulls_rolling_spool.png

Find the minimum coefficient of static friction between the cylinder and the horizontal surface that will allow for this perfect rolling to occur.

In the analysis we wrote down the equations that come from Newton's 2nd Law (for linear and rotational motion), and found the accelerations of the cylinder and block. The if the cylinder is just barely rolling because the coefficient of friction is as low as it can be, that means that the largest possible static friction is occurring:

\[f=\mu_sN = m_smg\nonumber\]

Putting this into the \(x\)-direction equation for the cylinder from Newton's 2nd Law, and plugging in the tension in the rope and the acceleration (both found in the analysis) gives:

\[f+T=ma\;\;\;\Rightarrow\;\;\;\mu_smg=-\frac{3}{4}ma+ma=\frac{1}{4}ma=\frac{1}{4}m\left(\frac{4}{11}g\right)\;\;\;\Rightarrow\;\;\;\mu_s=\frac{1}{11}\nonumber\]

Problem 5.10

Two ends of a massless rope are wound around two spools with equal masses and radii. One of the spools is a solid, uniform disk, while the other is a thin, hollow cylinder. The rope between them goes over a massless, frictionless pulley in a vertical plane. The spools are released from rest from the same height, and the rope does not slip over the pulley.

spools_around_pulley.png

The radius of the pulley is \(25cm\). Find the angle through which the pulley has turned when \(2.0s\) has elapsed.

From the result in the analysis , we can compute how much additional rope is in play. It is simply twice the distance that the spools fall:

\[\Delta L = 2\Delta y = 2\left(\frac{1}{2}at^2\right)=\left(\frac{3}{5}g\right)t^2=\frac{3}{5}\left(9.8\frac{m}{s^2}\right)\left(2.0s\right)^2=23.5m\nonumber\]

The torques are the same on both spools, but the solid spool has one-half the moment of inertia of the hollow one, so it has twice the angular acceleration, which means that over the same time period, it loses twice as much rope. Let's call the amount of rope lost by the hollow spool \(l\), so the amount lost by the solid spool is \(2l\). The rotation of the pulley makes up for this difference, which means it takes \(\frac{1}{2}l\) from the side where the solid spool is, and places it on the side of the hollow spool. Noting that the total rope lost by both spools combined is \(3l=\Delta L\), and using the no-slipping condition, we have:

\[s=R\theta\;\;\;\Rightarrow\;\;\; \theta = \frac{s}{R}=\frac{\frac{1}{2}l}{R}=\frac{\Delta L}{6R}=\frac{23.52m}{6\left(0.25m\right)}=15.7rad\nonumber\]

Problem 5.11

A board starts at rest and is free of any attachments (it is not pivoted on anything). It is pushed in opposite directions on both of its ends with forces of equal magnitude, at right angles to the board. The forces continue to be applied at right angles with the same magnitude, causing the board to rotate in the manner depicted in the diagram until the board has rotated by \(90^o\).

rotate_nonuniform_bar.png

The time it takes the board to rotate the \(90^o\) is \(t\). Derive an expression for the moment of inertia of this board in terms of \( t\), the length of board \(L\), and the force \(F\).

The analysis showed us that wherever the center of mass happens to be, the torque applied is still equal to \(FL\). Newton's 2nd Law for rotations gives:

\[\tau=I\alpha\;\;\;\Rightarrow\;\;\; I=\frac{\tau}{\alpha}=\frac{FL}{\alpha}\nonumber\]

The board starts from rest, and the torque remains constant, so it accelerates rotationally at a constant rate, which means its motion satisfies the usual kinematics equation:

\[\Delta\theta = \frac{1}{2}\alpha t^2\nonumber\]

We know that the rod rotates through an angle of \(\frac{\pi}{2}\) radians, and we know the time elapsed is \(t\), so we can solve for \(\alpha\) and this gives our final answer:

\[\alpha = \frac{2\Delta\theta}{t^2}=\frac{\pi}{t^2}\;\;\;\Rightarrow\;\;\;I=\frac{FLt^2}{\pi}\nonumber\]

Problem 5.12

The blob in the figure below is rigid and in static equilibrium. The two forces shown are two of the total of three forces exerted on the object.

blob_on_grid.png

The magnitude of \(F_1\) is three-quarters the magnitude of \(F_2\). Find the equation of the line along which the third force acts. 

For this object to remain in equilibrium, the net torque about every point in space must vanish. Consider the reference point \(x=-2\), \(y=+8\). Both of the given forces act through this point, so neither of them provides a torque around this reference point. There is only one more force present, and for the net torque around that reference point to be zero, the third force must also contribute zero torque, and this is only possible if that third force passes through the reference point.

In the analysis we found the tangent of the angle the force vector makes with the \(x\)-axis. This is the slope of the force line, and we are given the ratio of these two force magnitudes:

\[slope = \frac{F_2}{F_1} = \frac{4}{3}\nonumber\]

Now we know one point on the line and its slope, and these two quantities completely define it. Putting an arbitrary \(\left(x,y\right)\) point and the reference point into the slope equation gives us the equation of the line:

\[\frac{4}{3}=slope=\frac{y-y_{ref}}{x-x_{ref}}=\frac{y-8}{x+2}\;\;\;\Rightarrow\;\;\;y=\frac{4}{3}x+\frac{32}{3}\nonumber\]

Problem 5.13

Two painters carry a plank of plywood that they use for scaffolding over their heads on their way to the job site. The plank has a uniform mass distribution. Atop the plank is a can of paint weighing one third as much as the plank. The painter in the rear is holding the plank at the very end and the painter in front is holding the plank one quarter of the the plank length from the front. The can of paint is two-fifths of the plank length from the front. The plank is horizontal as they carry it.

painters_carry_plank.png

The can of paint weighs \(50N\). Find how much force each painter is exerting on the plank.

As is typical for these types of problems, the analysis often solves the whole thing. In this case, we know that the plank weighs three times as much as the paint, so the total weight carried is \(200N\). We found in the analysis how the load is distributed, so we have our answers already:

\[rear\;painter=0.30\left(200N\right)=60N\;,\;\;\;\;front\;painter=0.70\left(200N\right)=140N\nonumber\]

Problem 5.14

The diagram below depicts a yo-yo on an inclined plane with its string over a massless pulley and attached to a hanging block. The whole system is in static equilibrium.

yoyo_on_plane.png

The inner radius of the yo-yo is half the outer radius, and the coefficient of friction is 0.40.

  • Find the maximum angle \(\theta\) for which this system can be at static equilibrium (assume that the hanging mass can be adjusted to whatever is necessary).
  • The mass of the yo-yo is \(0.35kg\). If the angle \(\theta\) is at its maximum, find the hanging weight.

a. We have the equations of equilibrium from the analysis already. As the angle gets larger, the \(F_y\) equation shows that the normal force on the yo-yo by the plane gets smaller. This reduces the value of the maximum static friction force available. When the angle is so great that the new maximum friction force equals the actual friction force, then making the angle any larger would cause the yo-yo to slip. So the maximum angle without slippage means that \(f=\mu_sN\). Now we do some algebra.

Eliminate the tension first:

\[T=\frac{R}{r}f\;\;\;\Rightarrow\;\;\;\left(\frac{R}{r}-1\right)f-Mg\sin\theta=0\nonumber\]

Next eliminate the friction force:

\[f=\mu_sN\;\;\;\Rightarrow\;\;\;\left(\frac{R}{r}-1\right)\mu_sN-Mg\sin\theta=0\nonumber\]

And now the normal force:

\[N=Mg\cos\theta\;\;\;\Rightarrow\;\;\;\left(\frac{R}{r}-1\right)\mu_sMg\cos\theta-Mg\sin\theta=0\nonumber\]

Plugging-in the given \(R=2r\) and solving for \(\theta\) gives:

\[\theta = \tan^{-1}\mu_s=\tan^{-1}\left(0.4\right)=21.8^o\nonumber\]

b. Now to find the hanging mass, we need the tension. From the three equations above, we see that we can get the normal force from the mass of the yo-yo, the friction force from the normal force, and the tension from the friction:

\[mg=T=\frac{R}{r}f=\frac{R}{r}\mu_sN=\frac{R}{r}\mu_sMg\cos\theta\;\;\;\Rightarrow\;\;\;m=\left(2\right)\left(0.4\right)\left(0.35kg\right)\cos\left(21.8^o\right)\nonumber=0.26kg\]

See Our New JEE Book on Amazon

  • Kinematics of Rigid Bodies

kinematics of rigid bodies solved problems pdf

A rigid body is a system of particles in which the distance between the particles remains unchanged. It executes plane motion if all parts of the body moves in parallel planes.

Fixed Axis Rotation

In a fixed axis rotation, all lines in the body which are perpendicular to the axis of rotation rotates through the same angle in same time. Thus, all lines on a body in its plane of motion have the same angular displacement ($\theta$), the same angular velocity ($\omega=\mathrm{d}\theta/\mathrm{d}t$), and the same angular acceleration ($\alpha=\mathrm{d}\omega/\mathrm{d}t$).

The angular velocity and the angular displacement of a body rotating in a plane with a constant angular acceleration $\alpha$ are given by \begin{align} &\omega=\omega_0+\alpha t, \nonumber\\ &\theta=\theta_0+\omega_0 t+\tfrac{1}{2}\alpha t^2, \nonumber\\ &\omega^2=\omega_0^2+2\alpha t, \nonumber \end{align} where $\theta_0$ and $\omega_0$ are angular displacement and angular velocity at time $t=0$.

General Plane Rotation

In plane motion of a rigid body, the directions of angular velocity $\vec\omega$ and angular acceleration $\vec\alpha$ remains fixed (these are normal to the plane of rotation).

The velocities and accelerations of two points O and P on the rigid body are related by \begin{align} &\vec{v}_P=\vec{v}_O+\vec{\omega}\times{\vec{\mathrm{OP}}}=\vec{v}_O+\vec{\omega}\times\vec{r}, \nonumber\\ &\vec{a}_P=\vec{a}_O+\vec{\omega}\times\vec{\omega}\times\vec{r}+\vec{\alpha}\times\vec{r}, \nonumber \end{align} where $\vec{r}={\vec{\mathrm{OP}}}$ is the position vector from O to P. The acceleration term $\vec{\omega}\times\vec{\omega}\times\vec{r}$ is called centripetal acceleration and $\vec{\alpha}\times\vec{r}$ is called tangential acceleration .

If the axis of rotation is fixed then $\vec{v}_O=\vec{a}_O=\vec{0}$ for point O lying on the axis. This type of motion is called fixed axis rotation. In this case, the velocity of a point at a perpendicular distance $r$ from the axis is $v=\omega r$ (tangential). The centripetal acceleration of this point is $\omega^2 r$ and its tangential acceleration is $\alpha r$.

Instantaneous Axis of Rotation

An axis perpendicular to the plane of rotation that passes through the point of zero instantaneous velocity (this point may or may not lie on the body) is called instantaneous axis of rotation.

If two points on a rigid body are instantaneously at rest then all points on the line joining these points is at rest (instantaneous axis of rotation). The angular velocity of the body at that instant is along this line.

In-extensible String Constraint

The tangential component of velocity (i.e., component along the string) of any point on the string is same. Same is true for the acceleration.

Rolling without Slipping Constraint

Consider motion of a ball on a surface (fixed or moving). At the contact point, the ball may or may not slide on the surface. If tangential velocity of the ball at the contact point is equal to the velocity of the surface at this point then there is no slipping. Same is true for the acceleration.

Solved Problems from IIT JEE

  • A disc is rolling (without slipping) on a horizontal surface (IIT JEE 2004)
  • A sphere is rolling without slipping on a horizontal plane (IIT JEE 2009)
  • The figure shows a system consisting of (IIT JEE 2012)
  • Two identical discs of same radius R are rotating about their axes (IIT JEE 2012)
  • The general motion of a rigid body can be considered to be a combination of (IIT JEE 2012)
  • A Roller is Made by Joining Together Two Cones (JEE Mains 2016)

Related Topics

  • Moment of Inertia
  • Angular Momentum
  • Angular Momentum of a Projectile
  • Conservation of Angular Momentum
  • Fixed Axis of Rotation
  • Rolling without slipping of rings cylinders and spheres
  • Equilibrium of rigid bodies
  • Direction of frictional force on bicycle wheels
  • IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  • 300 Solved Problems on Rotational Mechanics by Jitender Singh and Shraddhesh Chaturvedi

JEE Physics Solved Problems in Mechanics

Book cover

Robot Mechanisms pp 1–60 Cite as

Kinematics of Rigid Bodies

  • Jadran Lenarčič 4 ,
  • Tadej Bajd 5 &
  • Michael M. Stanišić 6  

3364 Accesses

Part of the Intelligent Systems, Control and Automation: Science and Engineering book series (ISCA,volume 60)

The motion of rigid bodies is presented using standard vector and matrix algebra. Combinations of translations and rotations, as well as linear and angular velocities and linear and angular translations, are studied. The characteristic properties of the rotation matrix and of the homogeneous transformation matrix are described. Different ways to represent the orientation of the body are introduced, such as the Euler angles, the YPR angles and the invariants of the rotation matrix.

  • Angular Velocity
  • Planar Motion
  • Orientation Angle
  • Coordinate Frame
  • Rotation Matrix

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

This is a preview of subscription content, log in via an institution .

Buying options

  • Available as PDF
  • Read on any device
  • Instant download
  • Own it forever
  • Available as EPUB and PDF
  • Compact, lightweight edition
  • Dispatched in 3 to 5 business days
  • Free shipping worldwide - see info
  • Durable hardcover edition

Tax calculation will be finalised at checkout

Purchases are for personal use only

J. Angeles, Rational Kinematics (Springer, New York, 1988)

Book   MATH   Google Scholar  

O. Bottema, B. Roth, Theoretical Kinematics (Dover, New York, 1979/1990)

Google Scholar  

J.J. Craig, Introduction to Robotics: Mechanics and Control , 3rd edn. (Pearson/Prentice-Hall, Upper Saddle River, 2005)

J. Denavit, R.S. Hartenberg, J. Appl. Mech. 22 (2), 215 (1955)

MathSciNet   MATH   Google Scholar  

J. Lenarčič, in International Encyclopedia of Robotics , ed. by R. Dorf (Wiley, New York, 1988)

J.M. McCarthy, An Introduction to Theoretical Kinematics (MIT Press, Cambridge, 1990)

R. Paul, Robot Manipulators: Mathematics, Programming and Control (MIT Press, Cambridge, 1981)

L. Sciavicco, B. Siciliano, Modeling and Control of Robot Manipulators , 2nd edn. (Springer, London, 2000)

Book   Google Scholar  

L.W. Tsai, Robot Analysis, The Mechanics of Serial and Parallel Manipulators (Wiley, New York, 1999)

M. Vukobratović, M. Kirćanski, Kinematics and Trajectory Synthesis of Manipulation Robots (Springer, Berlin, 1986)

Download references

Author information

Authors and affiliations.

J. Stefan Institute, Ljubljana-Vic-Rudnik, Slovenia

Jadran Lenarčič

Faculty of Electrical Engineering, University of Ljubljana, Ljubljana-Vic-Rudnik, Slovenia

Aerospace and Mechanical Engineering, Notre Dame University, Notre Dame, IN, USA

Michael M. Stanišić

You can also search for this author in PubMed   Google Scholar

Rights and permissions

Reprints and permissions

Copyright information

© 2013 Springer Science+Business Media Dordrecht

About this chapter

Cite this chapter.

Lenarčič, J., Bajd, T., Stanišić, M.M. (2013). Kinematics of Rigid Bodies. In: Robot Mechanisms. Intelligent Systems, Control and Automation: Science and Engineering, vol 60. Springer, Dordrecht. https://doi.org/10.1007/978-94-007-4522-3_1

Download citation

DOI : https://doi.org/10.1007/978-94-007-4522-3_1

Publisher Name : Springer, Dordrecht

Print ISBN : 978-94-007-4521-6

Online ISBN : 978-94-007-4522-3

eBook Packages : Engineering Engineering (R0)

Share this chapter

Anyone you share the following link with will be able to read this content:

Sorry, a shareable link is not currently available for this article.

Provided by the Springer Nature SharedIt content-sharing initiative

  • Publish with us

Policies and ethics

  • Find a journal
  • Track your research

IMAGES

  1. Solved Problem 2: Kinematics of rigid bodies The wheel rolls

    kinematics of rigid bodies solved problems pdf

  2. Solved Problem I: Chapter 16 Kinematics of Rigid Bodies

    kinematics of rigid bodies solved problems pdf

  3. Dynamics: Practice #2 Kinematics & Kinetics of Rigid Bodies Kinematic

    kinematics of rigid bodies solved problems pdf

  4. (PDF) Chapter 15 " Kinematics of Rigid Bodies " Assignment 1

    kinematics of rigid bodies solved problems pdf

  5. SOLUTION: Statics of rigid bodies problems with solutions

    kinematics of rigid bodies solved problems pdf

  6. Module 5 Kinematics of Rigid Bodies Common Questions

    kinematics of rigid bodies solved problems pdf

VIDEO

  1. Lecture 21.2

  2. GATE-NPTEL

  3. Engineering Mechanics Lecture No 43a (Kinematics of Rigid Bodies Introduction)

  4. Kinematics of Rigid Bodies

  5. Vdo22 Geometry of Motion of Planar Rigid Bodies

  6. Kinematics of rigid bodies solved problem

COMMENTS

  1. PDF Chapter 15 KINEMATICS OF RIGID BODIES

    Considering the rotation of a rigid body about a fixed axis, the position of the body is defined by the angle q that the line BP, drawn from the axis of rotation to a point P of the body, forms with a fixed plane. The magnitude of the velocity of P is ds = dt = rq . is the time derivative of q. . sin f where Fig. 15.8 y

  2. PDF 6 Kinematics and kinetics of rigid bodies

    6 Kinematics and kinetics of rigid bodies 6.1 In-class A solid uniform cylinder of mass m and ra-dius R is placed on top of a xed cylinder of the same radius, and it is slightly tipped, as shown in the gure. Find the va-lue of the angle at which sliding begins as a function of the static friction coe cient . Solution 0. Notation

  3. PDF Plane Kinematics of Rigid Bodies

    Slide 1 A system of particles for which the distances between the particles remain unchanged. This is an ideal case. There is always some deformation in materials under the action of loads. This deformation can be neglected if the changes in the shape are small compared to the movement of the body as a whole. Particle Kinematics

  4. PDF Lecture 3: rigid body dynamics

    ry j Rigid Body Dynamics Linear Motion: F = ma = d(mv) dt sum of the forces is the time rate of change of linear momentum Works for particles - and also works for rigid bodies if the acceleration is at the center of mass! F =

  5. PDF ME 230 Kinematics and Dynamics

    The homework has usually 10-12 problems per week. Late ... Solutions to all problems solved in class will be ... Chapter 20 and 21: Three-Dimensional Kinematics of a Rigid Body & Overview of 3D Kinetics of a Rigid Body Chapter 22: Vibrations: under-damped free vibration, energy

  6. PDF Lecture 4: rigid body dynamics examples

    rigid body dynamics problems: 2D planar motion • Free Body Diagram! • 3 equations of motion: • problem constraints • mass moment of inertia calculation • can we solve? if not, need more eqns: • kinematics equations: connection between Fx = max Fy = may Mz = I↵ ↵, ! AND v, a Wednesday, April 17, 13

  7. PDF Rigid Body Equations of Motion

    Identify type of motion. Solve for linear and angular accelerations. Diagram. Assign inertial coordinate system. Draw complete free-body diagram. Draw kinetic diagram to clarify equations. Equations of motion. Apply 2 linear and 1 angular equations. Maintain consistent sense.

  8. PDF 8.01SC S22 Chapter 21: Rigid Body Dynamics About a Fixed Axis

    For the rigid body of mass m and momentum p = mV. cm , the translational equation of motion is still given by Eq. (21.2.1), which we repeat in the form. . Fext mA =. cm . (21.4.1) For fixed axis rotation, choose the z -axis as the axis of rotation that passes through the center of mass of the rigid body.

  9. PDF Rigid Body Kinematics

    Rigid Body Kinematics Master the conceptual, theoretical, and practical aspects of kinematics with this exhaust-ive text, which provides a rigorous analysis and description of general motion in mechanical systems, with numerous examples, from spinning tops to wheeled ground vehicles.

  10. PDF Chapter 6 Rigid Body Dynamics

    the properties of rigid bodies: the motion of a spinning top; a boomerang; the 'rattleback' and a Frisbee can all be explained using the equations derived in this section. Here is a quick outline of how we analyze motion of rigid bodies. 1. A rigid body is idealized as an infinite number of small particles, connected by two-force members. 2.

  11. 11: Rigid Body Kinematics

    11: Rigid Body Kinematics Rotational kinematics involving extended rigid bodies, as opposed to particles: fixed-axis rotation, gear- and belt-driven systems, relative and absolute motion analysis, and analysis using rotating …

  12. PDF Plane Kinetics of Rigid Bodies

    This pdf file contains the lecture notes of ME101 course on structural analysis, taught by Kaustubh Dasgupta at IIT Guwahati. The topic of this lecture is the analysis of statically indeterminate beams using the method of consistent deformations. The lecture explains the basic concepts, steps and examples of this method, as well as the advantages and limitations of it. The pdf file also ...

  13. PDF 6 Kinetics of rigid bodies

    necessary to solve for the motion of the rigid body.The rst equation is very similar to Newton's second law for a single particle, eq. (3.4). The mass of the entire rigid body multiplied by the acceleration of its center of mass equals the sum of all exter-nally applied forces. The rigid body can be replaced by a ctitious particle of mass

  14. Kinematics of Rigid Bodies Problems and Solutions

    Kinematics of Rigid Bodies Problems and Solutions | PDF Kinematics_of_Rigid_Bodies_problems_and_solutions - Read online for free.

  15. PDF Ch. 4: Plane Kinematics of Rigid Bodies

    Ch. 4: Plane Kinematics of Rigid Bodies 4.1 Introduction 4.1 Introduction Kinematics of rigid bodies involves both linear and angular quantities. Usage 1. In designing the machines to perform the desired motion. 2. To determine the motion resulting from the applied force. A rigid body A system of particles for which the distance

  16. LESSON 3. KINEMATICS OF A RIGID BODY SOLVED PROBLEMS

    KINEMATICS OF A RIGID BODY SOLVED PROBLEMS | Miguel Ángel Gisbert Soria - Academia.edu Download Free PDF LESSON 3. KINEMATICS OF A RIGID BODY SOLVED PROBLEMS Miguel Ángel Gisbert Soria See Full PDF Download PDF Related Papers Lecture notes on Kinematics sandeep bhaskar Download Free PDF View PDF Chapter 5 Plane Kinematics of Rigid Bodies 빵야빵야 모래반지

  17. (PDF) 11 kinematics of Rigid Bodies

    • Kinematics of rigid bodies: relations between time and the positions, velocities, and accelerations of the particles forming a rigid body. See Full PDF Download PDF Related Papers Kinematics Of Rigid Bodies Gwen Lore Crz Download Free PDF View PDF Chapter 5 Plane Kinematics of Rigid Bodies 빵야빵야 모래반지 Download Free PDF View PDF

  18. Dynamics of Rigid Bodies (Problems and Solutions)

    This excerpt, a glimpse into the forthcoming volume titled "Dynamics of Rigid Bodies, Problems and Solutions," is a testament to unwavering optimism. Crafted with the conviction that knowledge shared is a torch passed, this work aims to illuminate the intricate pathways of learning for students in the realm of engineering.

  19. Sample Problems

    Problem 5.3. A solid uniform sphere starts from rest and rolls down a flat ramp without slipping. The sphere descends a vertical distance of 3.6m by the time it reaches the bottom, and it takes 6.6s to make the journey. Find the angle that the ramp makes with the horizontal. Solution.

  20. PDF ME 2202 Dynamics of Rigid Bodies (Required)

    underlying kinematics and kinetics of rigid bodies in 2D and 3D motion. Objective 2: To educate students to identify, formulate and solve engineering problems in rigid body dynamics. 2.1 Students will demonstrate the ability to isolate rigid bodies and to draw clear and appropriate free body diagrams.

  21. Kinematics of Rigid Bodoes

    Kinematics of Rigid Bodies. A rigid body is a system of particles in which the distance between the particles remains unchanged. It executes plane motion if all parts of the body moves in parallel planes. A plane motion in which every line in the body remains parallel to its original position is called translation .

  22. Kinematics of Rigid Bodies

    The pose of a rigid body in space can be determined by use of so called natural coordinates. Natural coordinates are given by the position of three non-collinear points on the body P 1, P 2 and P 3. The three points are in Fig. 1.4 and are denoted by Cartesian vectors p 1, p 2 and p 3.

  23. Calculating Velocities, Angular Velocities, and Accelerations for Rigid

    Kinematics of Rigid Bodies Problems - Free download as PDF File (.pdf), Text File (.txt) or read online for free. intersting stuff