Dynamics of Rigid Bodies with Fixed Axis of Rotation

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In a fixed axis rotation, all particles of the rigid body moves in circular paths about the axis.

Consider a rigid body rotating about a fixed axis with an angular velocity $\omega$ and angular acceleration $\alpha$. Let $I$ be the moment of inertia about the axis of rotation.

The angular momentum of the rigid body about the fixed axis of rotation is given by \begin{align} L=I \omega \nonumber \end{align}

The torque on the rigid body about the axis of rotation is given by \begin{align} \tau=I \alpha. \nonumber \end{align}

Solved Problems from IIT JEE

Problem from JEE Mains 2017

A uniform disc of radius $R$ and mass $M$ is free to rotate about its axis. A string is wrapped over its rim and a body of mass $m$ is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is

a uniform disc of radius R and mass M
  1. $\frac{2mg}{2m+M}$
  2. $\frac{2Mg}{2m+M}$
  3. $\frac{2mg}{2M+m}$
  4. $\frac{2Mg}{2M+m}$

Solution: Let $T$ be tension in the string, $a$ be the acceleration of the body of mass $m$, and $\alpha$ be the angular acceleration of the disc of mass $M$ and radius $R$ (see figure).

a uniform disc of radius R and mass M

Apply Newton's second law on the body of mass $m$ to get \begin{align} \label{fjc:eqn:1} &mg-T=ma. \end{align} The forces on the disc are string tension $T$ at the contact point C, weight $Mg$ at the centre O and the reaction force at O. The torque about the point O is $\tau_O=TR$. The disc rotates about a fixed point O. Apply $\tau_O=I_O\alpha$ to get \begin{align} \label{fjc:eqn:2} TR=I_O\alpha=(MR^2/2)\alpha, \end{align} where $I_O$ is moment of inertia of the uniform solid disc about the axis of rotation. The acceleration of the string at the contact point C is $a$. The tangential acceleration of the pulley at the point C is $\alpha R$. These two accelerations should be equal for no slip at C i.e., \begin{align} \label{fjc:eqn:3} a=\alpha R. \end{align} Solve above equations to get \begin{align} a=\frac{2mg}{2m+M}.\nonumber \end{align} I encourage you to solve this problem by energy method.

Problem from IIT JEE 2002

Three particles A, B and C, each of mass $m$, are connected to each other by three massless rigid rods to form a rigid equilateral triangular body of side $l$. This body is placed on a horizontal frictionless table x-y plane) and is hinged to it at the point A, so that it can move without friction about the vertical axis through A (see figur). The body is set into rotational motion on the table about A with a constant angular velocity $\omega$.

  1. Find the magnitude of the horizontal force exerted by the hinge on the body.
  2. at time $T$, when the side BC is parallel to the x-axis, a force $F$ is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time $T$.
Three particles A, B and C

Solution: Consider the three masses and the connecting rods together as a system. The centre of mass of the system (G) is at a distance $\mathrm{AG}={l}/{\sqrt{3}}$ from the hinge point A.

Three particles A, B and C

The centre of mass moves in a circle with centripetal acceleration \begin{align} a_c=\omega^2 l/\sqrt{3}. \nonumber \end{align} The horizontal force acting on the system is the reaction at the hinge, $F_h$, which provides the necessary centripetal acceleration. Apply Newton's second law on the system to get \begin{align} F_h=(3m)a_c=\sqrt{3}m\omega^2 l. \nonumber \end{align} The force $F$ acting on B causes an anticlockwise torque $\tau=Fl\sqrt{3}/2$ about the point A. Since A is a fixed point, the torque $\tau$ is related to the angular acceleration $\alpha$ by \begin{align} Fl\sqrt{3}/2=I\alpha=2ml^2\alpha, \nonumber \end{align} which gives $\alpha={\sqrt{3}F}/{(4ml)}$. Total acceleration of the centre of mass immediately after a time $T$ is \begin{align} \label{jpb:eqn:4} \vec{a}&=a_x\,\hat\imath+a_y\,\hat\jmath \\ &= \alpha \frac{l}{\sqrt{3}}\,\hat\imath+a_c\,\hat\jmath \\ &= \frac{F}{4m}\,\hat\imath+\frac{\omega^2 l}{\sqrt{3}}\,\hat\jmath. \end{align} Let $\vec{F}_h=F_x\,\hat\imath+F_y\,\hat\jmath$ be the reaction force at the hinge and $\vec{F}_n=(F_x+F)\,\hat\imath+F_y\,\hat\jmath$, be the net force on the system. Newton's second law gives \begin{align} \label{jpb:eqn:5} \vec{F}_n&=(F+F_x)\,\hat\imath+F_y\,\hat\jmath \\ &=(3m)\vec{a}. \end{align} Substitute $\vec{a}$ from the previous equation into the last equation to get $F_x=-F/4$ and $F_y=\sqrt{3}m\omega^2l$.

Problem From AIEEE 2011

A pulley of radius 2 m is rotated about its axis by a force $F=(20t-5t^2)$ N (where, $t$ is time in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, then the number of rotations made by the pulley before its direction of motion is reversed is,

  1. more than 3 but less than 6
  2. more than 6 but less than 9
  3. more than 9
  4. less than 3

Solution: The torque on the pulley is \begin{align} \label{dic:eqn:1} \tau=rF=2(20t-5t^2)\,\mathrm{Nm}, \end{align} and its angular acceleration is \begin{align} \label{dic:eqn:2} \alpha&=\frac{\mathrm{d}\omega}{\mathrm{d}t} \\ &=\frac{\tau}{I}\\ &=(4t-t^2)\,\mathrm{rad/s^2}, \end{align} where $I=10$ kg-m2 is moment of inertia of the pulley. Integrate the above equation with initial condition $\omega=0$ to get the angular velocity \begin{align} \label{dic:eqn:3} \omega=(2t^2-t^3/3)\,\mathrm{rad/s}. \end{align} Integrate the above equation with initial condition $\theta=0$ to get the angular displacement \begin{align} \label{dic:eqn:4} \theta=(2t^2/3-t^4/12)\,\mathrm{rad}. \end{align}

A pulley of radius 2 m

The pulley comes to rest (momentarily) when $\omega=0$. Substitute $\omega=0$ in the expression for $\omega$ to get $t=6$ sec. Substitute $t=6$ sec in the expression for $\theta$ to get $\theta=36$ rad. Thus, number of rotations made by the pulley to come to rest are $n=\theta/(2\pi)=5.73$.

We should not ignore the fact that $\theta$ increases monotonically till pulley come to rest (see figure). This simplifies our calculations. We encourage you to find the number of rotations made by the pulley till $t=8$ sec.

Rotation of Raw and Boiled Eggs

If a raw egg and a boiled egg are spinned together with same angular velocity on the horizontal surface then which one will stops first?

Answer: Consider the rotation of hard boiled egg. The friction is the only reason which can stop it. The time to stop will depend on initial angular velocity and torque due to frictional force. Now, consider the raw egg. It is not a rigid body because fluid start rotating relative to the shell. It stops rotating because of (i) torque due to frictional force and (ii) loss of energy due to viscous drag. Going by this logic, raw egg should stop first if frictional forces are equal in two cases. It will be interesting to do the real experiment and get the result? We expect something strange when boiled egg is rotated very fast!

Related Topics

References

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. 300 Solved Problems on Rotational Mechanics by Jitender Singh and Shraddhesh Chaturvedi