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,
[tex]\begin{gathered} \tau=F\times r \\ =Fr\sin \theta \\ =mgr\sin \theta \end{gathered}[/tex]
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,
[tex]\begin{gathered} \tau=(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times\sin (90\degree) \\ =(1.5\text{ kg})\times(9.8\text{ m/s}^2)\times(20\text{ cm})\times(\frac{1\text{ m}}{100\text{ cm}})\times1 \\ =2.94\text{ N}\cdot m \end{gathered}[/tex]
In rotational dynamics torque is given as,
[tex]\tau=I\alpha[/tex]
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,
[tex]\alpha=\frac{\tau}{I}[/tex]
Substituting all known values,
[tex]\begin{gathered} \alpha=\frac{2.94\text{ N.m}}{2\text{ kg.m}^2} \\ =1.47\text{ rad/s}^2 \end{gathered}[/tex]
The angular velocity is given as,
[tex]\omega=\alpha t[/tex]
Here, t is the time.
Substituting all known values,
[tex]\begin{gathered} \omega=(1.47\text{ rad/s}^2)\times(4.2\text{ s}) \\ =6.174\text{ rad/s} \end{gathered}[/tex]
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,
[tex]\Theta=\omega t[/tex]
Substituting all known values,
[tex]\begin{gathered} \Theta=(6.174\text{ rad/s})\times(4.2\text{ s}) \\ =25.9308\text{ rad} \end{gathered}[/tex]
The number of revolutions of the
