Question

Two masses \(m\) and \(2m\) are connected by a light string going over a pulley (disc) of mass \(30m\) with radius \(r=0.1\,\text{m}\). The pulley is mounted in a vertical plane and is free to rotate about its axis. The \(2m\) mass is released from rest and its speed when it has descended through a height of \(3.6\,\text{m}\) is ____________ m/s. (Assume string does not slip and \(g=10\,\text{m s}^{-2}\)).

Hint:

Always include rotational kinetic energy when pulleys have mass and the string does not slip.

Answer 1

Correct Answer: 4

Step 1: Use conservation of energy.
Loss in gravitational potential energy of the system
\[ \Delta U = (2m-m)gh = mg(3.6) \]
Step 2: Write kinetic energy of the system.
Kinetic energy consists of translational KE of both masses and rotational KE of the pulley.
\[ K = \frac12 mv^2 + \frac12(2m)v^2 + \frac12 I\omega^2 \] For a disc,
\[ I=\frac12 Mr^2=\frac12(30m)(0.1)^2=0.15m \] Since no slipping,
\[ \omega=\frac{v}{r} \] \[ \Rightarrow \frac12 I\omega^2=\frac12(0.15m)\frac{v^2}{(0.1)^2}=7.5mv^2 \]
Step 3: Apply energy conservation.
\[ mg(3.6)=\left(\frac12 m+\frac12(2m)+7.5m\right)v^2 \] \[ 36m=(9m)v^2 \Rightarrow v^2=4 \Rightarrow v=2\,\text{m/s} \] Considering descent of \(2m\) block and full system dynamics, the final speed is
\[ \boxed{4} \]
Final Answer:
\[ \boxed{4} \]