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Two cylindrical hollow drums of radii R and 2R, and 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 separated by $(3R+\delta ).$ . They are now brought in contact $(\delta\rightarrow 0 ).$
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?

Answers (1)

$(a)v_{1}=\omega R$

$v_{2}=\omega \left (2R \right )=2R\omega$

The direction of

$v_{1}$ and $v_{2}$ at the point of contact is tangentially upward. The frictional force acts due to the difference in the two velocities.q

$f_{12}$ is force on 1 due to 2 and acts upward and $f_{21}$ acts downward

$f_{12}=-f_{21}$

(b) External forces acting on the system are $f_{12}$ and $f_{21}$ which are equal and opposite in direction

$f_{12}=-f_{21}$

$f_{12}+f_{21}=0$

$\left |f_{12} \right |=\left |f_{21} \right |=0$

external Torque = $F\times 3R$ (anti-clockwise)

As the Velocity of Drum 2 is twice,

$v_{2}=2v_{1}$

(c) When velocities of drum 1 and drum 2 become equal, no force of friction will act and hence

$v_{1}=v_{2}$

$\omega _{1}R=2\omega _{2}R$

$\frac{\omega _{1}}{\omega _{2}}=2$

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