Question

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of inertia of the pulley is 2 kg m². (i) Sketch the free body diagram of the 1.5 kg block.(ii) When the mass is released from rest, calculate the angular velocity and number of revolutions of the pulley at t = 4.2 s.

Answer 1

Part (i)

Free body diagram of the 1.5 kg block;

Part (ii)

Only 1 force is acting on the pulley is the weight of the block attached with the sting. The torque acting on the pulley is given as,

`\begin{gathered} \tau=F* r \\ =Fr\sin \theta \\ =mgr\sin \theta \end{gathered}`

Here, g is the acceleration due to gravity and the θ is the angle between force F and r (as force is acting tangentially hence θ=90°)

Substituting all known values,

`\begin{gathered} \tau=(1.5\text{ kg})*(9.8\text{ m/s}^2)*(20\text{ cm})*\sin (90\degree) \\ =(1.5\text{ kg})*(9.8\text{ m/s}^2)*(20\text{ cm})*(\frac{1\text{ m}}{100\text{ cm}})*1 \\ =2.94\text{ N}\cdot m \end{gathered}`

In rotational dynamics torque is given as,

`\tau=I\alpha`

Here, I is the moment of inertia of the pulley (I=2 kg.m²) and α is the angular acceleration.

The angular acceleration is given as,

`\alpha=(\tau)/(I)`

Substituting all known values,

`\begin{gathered} \alpha=\frac{2.94\text{ N.m}}{2\text{ kg.m}^2} \\ =1.47\text{ rad/s}^2 \end{gathered}`

The angular velocity is given as,

`\omega=\alpha t`

Here, t is the time.

Substituting all known values,

`\begin{gathered} \omega=(1.47\text{ rad/s}^2)*(4.2\text{ s}) \\ =6.174\text{ rad/s} \end{gathered}`

Therefore, the angular velocity of the pulley is 6.174 rad/s.

The angular displacement of the pulley in 4.2 s is given as,

`\Theta=\omega t`

Substituting all known values,

`\begin{gathered} \Theta=(6.174\text{ rad/s})*(4.2\text{ s}) \\ =25.9308\text{ rad} \end{gathered}`

The number of revolutions of the