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Two cylindrical hollow drums of radii $R$ and $2 R$, and of a common height $h$, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega^{\prime}$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by $3 R+\delta$. 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?
(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?
Solution:
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Verified Answer
Let us consider the figure shown below, that shown the frictional forces.
(a) The direction of $v_1$ and $v_2$ at point of contact are tangentially upward
So, $f^{\prime}=-f^{\prime \prime}, v_1=\omega R, v_2=\omega 2 R$.
(b) External forces acting on system are $f^{\prime}$ and $f^{\prime \prime}$ which are equal and opposite so net force acting on system
$$
\because f^{\prime}=-f^{\prime \prime} \text { or } f^{\prime}+f^{\prime \prime}=0, F_{\text {net }}=0
$$
$F^{\prime}=F=F^{\prime \prime}$ where $F^{\prime}$ and $F^{\prime \prime}$ are external forces through support.
So, External torque $=F \times 3 R$, (anti-clockwise)
As velocity of drum 2 is double i.e. $v_2=2 v_1$ as in part
(a).
(c) Let $\omega_1$ (anticlockwise) and $\omega_2$ (clockwise) be final angular velocities of smaller drum 1 and bigger drum 2 respectively.
Finally, when their velocities become equal, no force of friction will act due to no slipping or friction at this stage.
Hence, $v_1=v_2$
or $R \omega_1=2 R \omega_2 \Rightarrow\left(\frac{\omega_1}{\omega_2}=\frac{2}{1}\right)$
(a) The direction of $v_1$ and $v_2$ at point of contact are tangentially upward
So, $f^{\prime}=-f^{\prime \prime}, v_1=\omega R, v_2=\omega 2 R$.
(b) External forces acting on system are $f^{\prime}$ and $f^{\prime \prime}$ which are equal and opposite so net force acting on system
$$
\because f^{\prime}=-f^{\prime \prime} \text { or } f^{\prime}+f^{\prime \prime}=0, F_{\text {net }}=0
$$
$F^{\prime}=F=F^{\prime \prime}$ where $F^{\prime}$ and $F^{\prime \prime}$ are external forces through support.
So, External torque $=F \times 3 R$, (anti-clockwise)
As velocity of drum 2 is double i.e. $v_2=2 v_1$ as in part
(a).
(c) Let $\omega_1$ (anticlockwise) and $\omega_2$ (clockwise) be final angular velocities of smaller drum 1 and bigger drum 2 respectively.
Finally, when their velocities become equal, no force of friction will act due to no slipping or friction at this stage.
Hence, $v_1=v_2$
or $R \omega_1=2 R \omega_2 \Rightarrow\left(\frac{\omega_1}{\omega_2}=\frac{2}{1}\right)$
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