Question

1) Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separted by $3R+\delta$ They are now brought in contact $\delta \to 0$. (A) Show the frictional forces just after contact.

Hint: Friction force will act opposite to the relative motion of both drums.


2) Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities $\omega (anti-clockwise)$ and $\omega \left(clockwise\right)$, respectively. Their axes, fixed are parallel and in a horizontal plane separted by $3R+\delta$ They are now brought in contact $\delta \to 0$. (A) Show the frictional forces just after contact.

(B) Identify forces and torques external to the system just after contact.

(3) Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separted by $3R+\delta$ They are now brought in contact $\delta \to 0$. (A) Show the frictional forces just after contact.

(C) What would be the ratio of final angular velocities when friction ceases?

Answer 1

(A) The direction of rotation of both drum are in opposite sense therefore friction will act opposite to their relative motion. Here angular velocity of small drum is anti-clockwise and for large drum is clockwise.

At the point of contact their tangential velocity are in same direction upward. From the relative motion the final tangential velocity of smallest drum is in downward direction so friction will act in upward direction $\left(f_1\right)$

From the Newton's third law downward friction will act $\left(f_2\right)$
on larger drum.



Final Answer:

For smallar drum fricton force $\left(f_1\right)$ will be upward.

For larger drum fricton force $\left(f_2\right)$ will be downward.

(B) Step 1: Draw a rough diagram.

Step 2: Apply torque about any point if net force is zero.

Formula used: $\tau =\overrightarrow{r}\times \overrightarrow{F}$

There is no translational motion to the point of contact so equal and opposite force $\left(F^{\prime }and{F}^{\prime \prime }\right)$ will work on both drums. If two forces are in opposite sense then it said to be net force is zero, here $F^{\prime }and{F}^{\prime \prime }$ are working in opposite sense. It seems ${\overrightarrow{F}}_{net}=0$ on both drums.

Here we could apply external torque.

${\tau }_{ext}=\overrightarrow{r}\times \overrightarrow{F}=rFsin\theta$

$\because \theta ={90}^{\circ },F^{\prime }=F^{\prime \prime }=F\left(\text{equal and opposite}\right)$

Here, $r=3r+\delta \left(\text{distance between two drums from smallest drum}\right)$

As $(\delta \to 0).$

So $r=3R$

${\tau }_{ext}=F\times 3R\left(anticlockwise\right)$

Final Answer : External torque $=F\times 3R,anticlockwise$

c) Formula used: $v=R\omega$

Let ${\omega }_{1}and{\omega }_2$ be final angular velocities (anticlockwise and clockwise respectively).

Velocity of first drum $v_1=R{\omega }_1$

Velocity of first drum $v_2=2R{\omega }_2$

There will be no friction. So, at a contact point their translational velocities are same.

$v_1=v_2$

$R{\omega }_1=2R{\omega }_2$

$\frac{{\omega }_1}{{\omega }_2}=2$

Final Answer: 2