# 19.11 Work and energy in rolling motion  (Page 3/3)

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$\begin{array}{l}⇒K=\left(I+M{R}^{2}\right)x\frac{{\omega }^{2}}{2}\end{array}$

## Distribution of kinetic energy

The rolling motion is characterized by unique distribution of kinetic energy between translation and rotation. This distribution is dependent on the distribution of mass about the axis of rotation. If “T” and “R” subscripts denote translational and rotational motions respectively, then :

$\begin{array}{l}\frac{{K}_{T}}{{K}_{R}}=\frac{M{{V}_{C}}^{2}}{I{\omega }^{2}}=\frac{M{{V}_{C}}^{2}{R}^{2}}{I{{V}_{C}}^{2}}\end{array}$

$\begin{array}{l}\frac{{K}_{T}}{{K}_{R}}=\frac{M{R}^{2}}{I}\end{array}$

For a ring,

$\begin{array}{l}I=M{R}^{2}\end{array}$

$\begin{array}{l}⇒\frac{{K}_{T}}{{K}_{R}}=\frac{M{R}^{2}}{M{R}^{2}}=1\end{array}$

This means that energy is equally distributed between translation and rotation in the case of ring. Similarly, for a solid sphere,

$\begin{array}{l}⇒\frac{{K}_{T}}{{K}_{R}}=\frac{5M{R}^{2}}{2M{R}^{2}}=\frac{5}{2}\end{array}$

We can infer from the values of ratio in two cases that the ratio of kinetic energies in translation and rotation is equal to the inverse of the numerical coefficient of MI of the rolling body.

## Work – kinetic energy theorem

Work by net external force is related to kinetic energy of the rolling body in accordance with work – energy theorem as :

$\begin{array}{l}⇒{W}_{\mathrm{ext}}=\Delta K\end{array}$

In rolling, the important feature of work by net external force on the left of the equation is that this net external force term does not include friction as work by friction is zero. This is the first difference between consideration of work – energy theorem in pure rolling as against in pure translation or pure rotation. Secondly, the change in kinetic energy in rolling refers to both translational and rotational kinetic energy.

## Conservation of mechanical energy in rolling

The work by friction in rolling is zero. The presence of friction does not result in dissipation of energy like heat to the system and its surrounding. It means that total mechanical energy comprising of potential and kinetic energy of a rolling body system remains same or is conserved.

This is contrary to the motion of translation alone i.e. sliding - in which case, friction converts some of the mechanical energy into heat, sound etc, which can not be regained by the system as mechanical energy. Thus, mechanical energy of a body in translation only (sliding) is not conserved. In the case of pure rotation, mechanical energy of the rotating body is not conserved as the force of friction at the shaft (axis of rotation) results in dissipation of energy.

Conservation of mechanical energy is the characterizing feature of pure rolling. This is significant as mechanical energy is conserved even when friction is present.

Problem : A small spherical ball of mass “m” and radius “r” rolls down a hemispherical shell of radius “R” as shown in the figure. The ball is released from the top position of the shell. What is the angular velocity of the spherical ball at the bottom of the shell as measured about perpendicular axis through the center of hemispherical shell.

Solution : Let the ball reaches the bottom with certain velocity “ ${v}_{C}$ ”. Then, the angular velocity of the ball about perpendicular axis through the center of hemispherical shell is :

$\begin{array}{l}{\omega }_{0}=\frac{{v}_{C}}{\left(R-r\right)}\end{array}$

Note that the linear distance between the center of ball and center of hemispherical shell is (R-r). In order to evaluate this equation, we need to find linear velocity of the spherical ball. Now, the ball is rolling along a curved path, while transcending a height of (R-r).

According to conservation of mechanical energy,

$\begin{array}{l}K=\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}M{{V}_{C}}^{2}+\frac{1}{2}xI{\omega }^{2}=Mg\left(R-r\right)\end{array}$

Using, $\omega =\frac{{V}_{C}}{R}$ and putting expression of MI of the sphere,

$\begin{array}{l}⇒\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}M{{V}_{C}}^{2}+\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\frac{2M{R}^{2}}{5}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\frac{{{V}_{C}}^{2}}{{R}^{2}}=Mg\left(R-r\right)\end{array}$

$\begin{array}{l}⇒\frac{1}{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}M{{V}_{C}}^{2}+\frac{1}{5}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}M{{V}_{C}}^{2}=Mg\left(R-r\right)\end{array}$

$\begin{array}{l}⇒{V}_{C}=\surd \left\{\frac{10g\left(R-r\right)}{7}\right\}\end{array}$

The required angular velocity about perpendicular axis through the center of hemispherical shell is obtained by substituting for vC in equation – 17 :

$\begin{array}{l}⇒{\omega }_{0}=\frac{{v}_{C}}{\left(R-r\right)}=\surd \left\{\frac{10g}{7\left(R-r\right)}\right\}\end{array}$

## Summary

1: Work by friction is zero in rolling.

2: Work by friction in translation and rotation are equal in magnitude but opposite in direction.

3: Gravitational potential energy change in rolling motion is due to change in position of the COM due to translation – not rotation.

4: The rolling has both kinetic energies corresponding to translation and rotation. The distribution of kinetic energy between two motions depend on the distribution of moment of inertia as :

$\begin{array}{l}\frac{{K}_{T}}{{K}_{T}}=\frac{M{R}^{2}}{I}\end{array}$

5: Conservation of mechanical energy is the characterizing feature of pure rolling. This is significant as mechanical energy is conserved even when friction is present.

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