Question

A wheel of moment of inertia $ 2.5\text{ }kg-{{m}^{2}} $ has an initial angular velocity of $ 40\text{ }rad\text{ }{{s}^{-1}} $ . A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to $ 60\text{ }rad\text{ }{{s}^{-1}} $ is

• A. 4s
• B. 6s
• C. 5s
• D. 2.5s
• E. 4.5s

Answer 1

Answer: (C)

Given: $ MI=2.5\text{ }kg\text{ }{{m}^{-2}} $
$ \omega =40\text{ }rad\text{ }{{s}^{-1}} $
$ \tau =10\text{ }Nm $
As $ \tau =I\alpha $
$ \therefore $ $ 10=2.5\alpha $
$ \alpha =4\text{ }rad\text{ }{{s}^{-2}} $
Now, $ \omega ={{\omega }_{0}}+\alpha t $
$ \therefore $ $ 60=40+4\times t $ or
$ 20=4t $
$ t=5s $