Question

A wheel has moment of inertia `5 xx 10^(-3) kg m^(2)` and is making `20 "rev" s^(-1)`. The torque needed to stop it in `10 s` is….. `xx 10^(-2) N-m`
A. `2 pi`
B. `2.5 pi `
C. `4 pi`
D. `4.5 pi`

Answer 1

Correct Answer - A
Here, `I = 5 xx 10^(-3) kg m^(2), n^(1) = 20 rps`
`tau = ?, n_(2) = 0, t = 10 s`
`tau = I alpha = (I(omega_(2) - omega_(1)))/(t) = (I xx 2pi (n_(2) - n_(1)))/(t)`
`= (5 xx 10^(-3) xx 2pi (0 - 20))/(10)`
`= 2pi xx 10^(-2) N-m`