Question

This bar starts its rotational motion with an angular velocity of 10 rd/s at t=2s and ends it at t=4s with an angular velocity of 42 rd/s. ( Given The final moment of inertia "I" of the system equals: I = 8 kg.m2) Its angular acceleration and torque are respectively :

Answer 1

a) We can find the angular acceleration of the bar as follows:

$$\omega=\omega_0+\alpha t,$$ $$\alpha=\dfrac{\omega-\omega_0}{t},$$ $$\alpha=\dfrac{42\ \dfrac{rad}{s}-10\ \dfrac{rad}{s}}{2\ s}=16\ \dfrac{rad}{s^2}.$$

b) We can find the torque as follows:

$$\tau=I\alpha=8\ kg\cdot m^2\cdot16\ \dfrac{rad}{s^2}=128\ N\cdot m.$$