Question

Figure shows two cylinders of radii $r_{1}and{r}_2$ having moments of inertia $I_{1}and{I}_2$ about their respective axes. Initially, the cylinders rotate about their axes with angular speeds ${\omega }_{1}and{\omega }_2$ as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinder 1 and 2 respectively after the slipping ceases.

• A. $\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_2,\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_2^2+I_1{r}_1^2}\right]r_2$
• B. $\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_2,\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_1$
• C. $\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_1{r}_1^2+I_2{r}_2^2}\right]r_2,\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_1{r}_1^2+I_2{r}_2^2}\right]r_1$
• D. $\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_2,\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_2^2+I_1{r}_1^2}\right]r_1$

Answer 1

The correct option is B

$\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_2,\left[\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right]r_1$

When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If ${\omega }^{\prime }1and{\omega }^{\prime }2$ be the respective angular speeds, we have ${\omega }^{\prime }1r_1={\omega }^{\prime }2r_2$...(i)

The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is fr1 and that on the second is $f{r}_2$. Assuming ${\omega }_1{r}_1>{\omega }_2{r}_2$, the corresponding angular impulses are -$f{r}_{1}tandf{r}_{2}t$. We, therefore, have -$f{r}_{1}t=I_1({\omega }^{\prime }1-{\omega }_1)$

$f{r}_{1}t=I_1({\omega }^{\prime }1-{\omega }_1)$

and $f{r}_{2}t=I_2({\omega }^{\prime }2-{\omega }_2)$

or,$-\frac{I_1}{r_1}({\omega }_1^{\prime }-{\omega }_1)\frac{I_2}{r_2}({\omega }_2^{\prime }-{\omega }_2)$ ...(ii)

solving (i) and (ii),

${\omega }_1^{\prime }=\left(\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right)r_{2}and{\omega }_2^{\prime }=\left(\frac{I_1{\omega }_1{r}_2+I_2{\omega }_2{r}_1}{I_2{r}_1^2+I_1{r}_2^2}\right)r_1$.