Question

A constant torque of 18.1 N · m is applied to a grindstone whose moment of inertia is 0.131 kg · m2 . Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 15.2 rev. Answer in units of rad/

Answer 1

162.45 rad/s

Step-by-step explanation:

We are given that

Torque,
`\tau=18.1 N`

Moment of inertia,
`I=0.131 kgm^2`

Initial angular velocity,
`\omega_0=0`

Angular displacement,
`\theta=15.2 rev=15.2(2\pi)`rad

1 rev=
`2\pi` rad

According to Work energy theorem

Work done=Change in kinetic energy

`\tau\theta=(1)/(2)I\omega^2-(1)/(2)I\omega^2_0`

Substitute the values

`18.1* 15.2(2\pi)=(1)/(2)(0.131)\omega^2-0`

`\omega^2=(18.1* 15.2(2\pi)* 2)/(0.131)`

`\omega=\sqrt{(18.1* 15.2(2\pi)* 2)/(0.131)}`

`\omega=162.45 rad/s`