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