Question

A point mass of 2kg is moving around a circle of 1 m. If it takes 9 Nm to stop the particle in 1s, how many Nm would it take to stop the particle in half the time?

Answer 1

here,

mass of object , m = 2 kg

radius , r = 1 m

the angular acceleration , alpha1 = T /I

alpha1 = 9 /(m * r^2)

alpha1 = 9 /(2 * 1^2) = 4.5 rad/s^2

let the initial angular velocity be w0

using first equation of motion

0 = w0 + alpha1 * t1

0 = w0 - 4.5 * 1

w0 = 4.5 rad/s

when the time taken , t2 = t1 /2 = 0.5 s

the angular acceleration required , alpha2 = w0 /t2

alpha2 = 4.5 /(1/2) = 9 rad/s^2

the torque required , T2 = I * alpha2

T2 = (2 * 1^2) * 9 N.m

T2 = 18 N.m