Question

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular acceleration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.

Answer 1

The magnitude of the applied torque is
`6.0*10^(-2)\ N-m`

(e) is correct option.

Step-by-step explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

`\tau=I*\alpha`

Here, I = mk²

`\tau=mk^2*\alpha`

Put the value into the formula

`\tau=3*(0.2)^2*0.5`

`\tau=0.06\ N-m`

`\tau=6.0*10^(-2)\ N-m`

Hence, The magnitude of the applied torque is
`6.0*10^(-2)\ N-m`