Question

A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 s $\text{rad}{\text{s}}^{-1}$ . If the mass is brought to rest in 10 s by a brake, what is the magnitude of the torque applied?

• A. 0.5 N-m
• B. 0.9 N-m
• C. 1.2N-m
• D. 2.3 N-m

Answer 1

Answer: (B)

Given, ${\omega }_1=10\text{rad}{\text{s}}^{-1},$ ${\omega }_2=0,$ $t=10s$ $\therefore$ $\alpha =\frac{{\omega }_2-{\omega }_1}{t}=\frac{0-10}{10}$ $=-1rad{s}^{-2}$ (Negative sign shows retardation) Now, moment of inertia $I=m{r}^2,$ $=10\times {(0.3)}^2=0.9kg\text{-}{m}^2$ $\therefore$ Torque, $\tau =I\alpha =0.9\times 1=0.9N\text{-}m$