Rotational Kinetic Energy - Problem Solving

A variety of problems can be framed on the concept of rotational kinetic energy. The problems can involve the following concepts,

1) Kinetic energy of rigid body under pure translation or pure rotation or in general plane motion.
2) Work done by torque and its relation with rotational kinetic energy in case of fixed axis rotation.
3) Conservation of mechanical energy.

A rod of mass 'M' and length 'L' is rotating about an axis passing through its end and perpendicular to its length. If the angular velocity of rotation at an instant is \(\omega \) then find its kinetic energy.

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As the axis of rotation of the rod is fixed thus the rod is in pure rotation and its rotational kinetic energy is given by
\[KE = \frac{1}{2}{I_{rot}}{\omega ^2}\] here, \({I_{rot}}\) is the moment of inertia of rod about the axis of rotation, which is \[{I_{rod\,about\,end}} = \frac{{M{L^2}}}{3}\] Thus the kinetic energy is given by
\[KE = \frac{1}{2}\frac{{M{L^2}}}{3}{\omega ^2} = \frac{{M{L^2}{\omega ^2}}}{6}\]

A uniform hoop (ring) of mass M and radius R is rolling without slipping on a horizontal ground with its center having velocity 'v'. Find the kinetic energy of the hoop.

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The ring is in general plane motion, thus its motion can be thought as the combination of pure translation of the center of mass and pure rotation about the center of mass.
The kinetic energy of the hoop will be written as, \[KE = \frac{1}{2}MV_{_{cm}}^2 + \frac{1}{2}{I_{cm}}{\omega ^2}\]

Here \({V_{cm}}\) is the speed of the center of mass and \({I_{cm}}\) is the moment of inertia about an axis passing through its center of mass and perpendicular to the plane of the hoop. \[{I_{cm,hoop}} = M{R^2}\] \[KE = \frac{1}{2}Mv_{}^2 + \frac{1}{2}M{R^2}{\omega ^2}\] For pure rolling motion (rolling without slipping)
\[v = r\omega \] \[KE = \frac{1}{2}Mv_{}^2 + \frac{1}{2}M{v^2} = M{v^2}\]

A pulley of mass M has a thread wound around it tightly, as shown in the diagram. A constant force starts acting on the open end of the thread. If the pulley is initially at rest, then find the angular speed of the pulley as a function of angle rotated by the pulley.

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There are three forces acting on the pulley
1) Force by thread
2) Gravitational force acting on the center of mass of the pulley
3) Force by hinge

Torque of hinge force and gravitational force about the center of the pulley is zero as they pass through the center itself.

Thus, the net torque about the center of the pulley equals \[\vec \tau = \vec r \times \vec F\]
\[\tau = FR\]

The torque is constant, thus the net work done by the torque on rotating the pulley by an angle \(\theta \) equals, \[W = FR\theta \] The work done by the torque goes into increasing the rotational kinetic energy of the pulley, Thus, according to the work energy theorem for rotation,
\[\begin{array}{l} {W_\tau } = \Delta KE\\ FR\theta = \frac{1}{2}I({\omega ^2} - {0^2}) \end{array}\] A pulley can be considered as a disc, thus the moment of inertia \(I = \frac{{M{R^2}}}{2}\) \[\omega = \sqrt {\frac{{4F\theta }}{{MR}}} \]

A sphere is released from the top of a rough inclined plane. The friction is sufficient so that the sphere rolls without slipping. Mass of the sphere is M and radius is R. The height of the center of the sphere from ground is h. Find the speed of the center of the sphere as it reaches the bottom of the sphere.

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In case of pure rolling on the fixed inclined plane, the point of contact remains at rest and work done by friction is zero. If sphere and earth are taken into one system, then the gravitational force becomes internal force. Other external force, Normal reaction is perpendicular to the direction of motion, thus will not do any work. Thus, no external force or non conservative forces are doing work, and mechanical energy of the system can be conserved.

When the ball reaches the bottom of the inclined plane, then its center is moving with speed 'v' and the ball is also rotating about its center of mass with angular velocity \(\omega \). In pure rolling motion, v and \(\omega \) are related as
\[v = R\omega \] As the ball comes down the potential energy decreases and therefore kinetic energy increases. The center of ball decends by 'h-R',
Loss in potential energy = gain in kinetic energy
\[Mg(h - R) = \frac{1}{2}M{v^2} + \frac{1}{2}{I_{cm}}{\omega ^2}\] Moment of inertia of sphere about an axis passing through the center of mass equals \[{I_{cm,sphere}} = \frac{2}{5}M{R^2}\] Therefore, \[\begin{array}{l} g(h - R) = \frac{7}{{10}}{v^2}\\ v = \sqrt {\frac{{10g(h - R)}}{7}} \end{array}\]