19.7 Rolling with sliding  (Page 3/3)

$$\begin{array}{l}a=\frac{f}{M}\end{array}$$

Substituting in equation of motion (equation - 4),

Let us, now, analyze rotational motion. Here,

$$\begin{array}{l}\omega =-{\omega }_0+\alpha t\end{array}$$

Initial angular velocity is clockwise and hence negative. The rotational acceleration is obtained from the Newton’s law in angular form as torque divided by moment of inertia (angular acceleration is positive as it anti-clockwise) :

$$\begin{array}{l}\alpha =\frac{\tau }{I}=\frac{fR}{I}\end{array}$$

Torque and angular acceleration due to friction are anti-clockwise and hence are positive. Substituting in the equation of angular motion,

Thus, we have two equations : one for translational motion and another for rotational motion. We need to eliminate time “t” from these equations. From equation of translational motion (equation – 5), we have :

$$\begin{array}{l}t=\frac{vM}{f}\end{array}$$

Substituting the value of time in the equation of rotational motion, we have :

$$\begin{array}{l}\omega =-{\omega }_0+\frac{vMR}{I}\end{array}$$

Now, we use the condition of rolling, v = -ωR as "v" and ω have opposite signs. In this case, we multiply the equation for ω by “-R” to get the equation for translational velocity :

Substituting the expression of MI of hollow sphere about a diameter,

$$\begin{array}{l}\Rightarrow v={\omega }_{0}R-\frac{3vM{R}^2}{2M{R}^2}\\ \Rightarrow v={\omega }_{0}R-\frac{3v}{2}\\ \Rightarrow \frac{5v}{2}={\omega }_{0}R\\ \Rightarrow v=\frac{2{\omega }_{0}R}{5}\end{array}$$

Angular velocity is given by :

$$\begin{array}{l}\Rightarrow \omega =-\frac{v}{R}=-\frac{\frac{2{\omega }_{0}R}{5}}{R}=-\frac{2{\omega }_0}{5}\end{array}$$

The rigid body initially having greater linear velocity.

Problem : A disk of mass “M” and radius “R” moves on a horizontal surface with translational speed “ $v_0$ ” and angular speed “ $\frac{v_0}{3R}$ ”. Find linear velocity, when the disk starts moving with pure rolling.

Solution : According to the question, the initial motion is not pure rolling. For pure rolling, translational speed is linked to angular speed by the equation,

$$\begin{array}{l}v=\omega R\end{array}$$

But here, translational velocity ( $v_0$ ) is greater than the required angular velocity for rolling. The linear velocity corresponding to the given angular velocity is only " $\omega R=\frac{v_0}{3R}xR=\frac{v_0}{3}$ ". This means that the disk translates more than "2πR" in one revolution. This, in turn, means that the disk also slides on the surface, besides rotating.

The sliding motion is opposed by friction acting in opposite direction to the motion. Now as the disk decelerates, let “a” be the deceleration. The deceleration, “a”, is given by Newton’s second law. Here, force responsible for deceleration is kinetic friction.

$$\begin{array}{l}a=-\frac{F_K}{M}\end{array}$$

Negative sign shows that it acts in the direction opposite to x-axis. The velocity of center of mass, “v”, at a time “t” is given by the equation of linear motion for constant acceleration (note that we have not used the subscript “C” for center of mass to keep notation simple) :

$$\begin{array}{l}v=v_0+at\end{array}$$

Substituting the expression of acceleration,

We can similarly find expression for angular velocity at a given time. First, we need to calculate angular acceleration resulting from the torque applied by the friction. Note here that friction causes angular acceleration – not deceleration.

Now, angular acceleration,”α”, is given as :

$$\begin{array}{l}\alpha =\frac{\tau }{I}=-\frac{2F_{K}R}{M{R}^2}=-\frac{2F_K}{MR}\end{array}$$

Angular acceleration is clockwise and hence negative. The angular velocity, “ω”, at a time “t” is given by :

$$\begin{array}{l}\omega =-{\omega }_0+\alpha t\end{array}$$

Substituting expression of angular aceleration,

$$\begin{array}{l}\omega =-{\omega }_0-\frac{2F_K}{MR}xt\end{array}$$

According to equation of rolling (negative sign for relation of velocities as against unsigned relation for speeds),

We have two expressions (equations 8 and 10) for translation velocity at time “t”. Now, "t" is obtained from equation - 8 as :

$$\begin{array}{l}t=\frac{(v_0-v)M}{2F_K}\end{array}$$

Substituting expression of “t” in the equation - 10, we have :

$$\begin{array}{l}v=\frac{v_0}{3}+2(v_0-v)\\ \Rightarrow 3v=\frac{7v_0}{3}\\ \Rightarrow v=\frac{7v_0}{9}\end{array}$$

Acknowledgement

Author wishes to place special thanks to Wei Yu for their valuable suggestions on the subject discussed in this module.