Question

Two cylinders with radii $r_1$ and $r_2$ and rotational inertia $I_1$ and $I_2$ are supported on their horizontal axles. The first one is set in rotation with angular velocity $\omega$. The axle of the other cylinder (smaller) is moved until it touches the large cylinder and is caused to rotate by the frictional forces between the two. Find the angular velocity of the two cylinders when slipping ceases finally because of the frictional forces between them.

• A. ${\omega }_1=\frac{I_1{r}_2^2\omega }{I_1{r}_2^2+I_2{r}_1^2}$
• B. ${\omega }_1=\frac{I_1{r}_2^2\omega }{I_1{r}_2^2-I_2{r}_1^2}$
• C. ${\omega }_2=\frac{I_1\omega r_1}{I_1{r}_2-I_2{r}_1}$
• D. ${\omega }_2=\frac{I_1\omega r_2}{I_1{r}_2-I_2{r}_1}$

Answer 1

The correct option is A ${\omega }_1=\frac{I_1{r}_2^2\omega }{I_1{r}_2^2+I_2{r}_1^2}$

In this case, due to presence of forces at axle some unbalanced torque will always exist on system hence principle of conservation of angular momentum can not be applied.

When we move the axles and the cylinders touch each other, then frictional force $\left(f\right)$ acts on the cylinder $\left(1\right)$ and cylinder $\left(2\right)$ as shown in figure.


Due to the frictional force$\left(f\right)$, cylinder $\left(1\right)$ and cylinder $\left(2\right)$ attain angular velocity ${\omega }_1$ & ${\omega }_2$ at the instant when slipping between them ceases.

Hence, velocity of cylinders at contact point will be same.
$v_1=v_2$
$r_1{\omega }_1=r_2{\omega }_2$
${\omega }_2=\frac{r_1{\omega }_1}{r_2}----\left(i\right)$
Applying the equation of angular impulse for both cylinders, considering the anticlockwise sense of rotation as $+ve$ direction.
$\overrightarrow{\tau }={\overrightarrow{L}}_f-{\overrightarrow{L}}_i$
$\Rightarrow \int f{r}_{1}dt=-I_1{\omega }_1-(-I_1\omega )$
$\Rightarrow \int f{r}_{1}dt=-I_1{\omega }_1+I_1\omega ....\left(ii\right)$
Similarly for cylinder $2$,
$\Rightarrow \int f{r}_{2}dt=I_2{\omega }_2-0=I_2{\omega }_2....\left(iii\right)$
On dividing Eq.$\left(ii\right)\&\left(iii\right)$,
$\Rightarrow \frac{-I_1{\omega }_1+I_1\omega }{I_2{\omega }_2}=\frac{r_1}{r_2}$
$\Rightarrow -I_1{\omega }_1+I_1\omega =\frac{I_2{\omega }_2{r}_1}{r_2}$
$\Rightarrow -I_1{\omega }_1{r}_2+I_1\omega r_2=I_2{\omega }_2{r}_1$
Substituting ${\omega }_2=\frac{{\omega }_1{r}_1}{r_2}$ in above equation we get,
$\Rightarrow -I_1{\omega }_1{r}_2^2+I_1\omega r_2^2=I_2{r}_1^2{\omega }_1$
$\Rightarrow {\omega }_1(I_1{r}_2^2+I_2{r}_1^2)=I_1\omega r_2^2$
$\therefore {\omega }_1=\frac{I_1\omega r_2^2}{I_1{r}_2^2+I_2{r}_1^2}$

$\&{\omega }_2=\frac{{\omega }_1{r}_1}{r_2}$
$\Rightarrow {\omega }_2=\frac{I_1\omega r_1{r}_2}{I_1{r}_2^2+I_2{r}_1^2}$
$\therefore \left(a\right)$ is the only correct option.