Question

The diameter of a solid disc is 0.5 m and its mass is 16 Kg. Its angular velocity increases from zero to 120 rotation/minute in 8 seconds, at what rate is work done by the torque at the end of 8th second?

• A. $\pi Watt$
• B. ${\pi }^{2}Watt$
• C. ${\pi }^{3}Watt$
• D. ${\pi }^{4}Watt$

Answer 1

The correct option is B ${\pi }^{2}Watt$

$\omega =\frac{{\theta }_f-{\theta }_i}{t-t_0}=\frac{120\times 2\pi -0}{60}rad/s=4\pi rad/s$
Acceleration at the end of the 8th second will be: $\alpha =\frac{4\pi -0}{8}=\frac{\pi }{2}rad/s^2$
Power is calculated as: $\tau .\omega =I\alpha \omega =\frac{16\left({0.25}^2\right)}{2}\frac{\pi }{2}.\left(4\pi \right)$ $=\frac{16\left({0.25}^2\right)}{2}\frac{\pi }{2}.\left(4\pi \right)={\pi }^{2}Watts$