Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `4 m`. If the object's angular velocity changes from `1 Hz` to `6 Hz` in `3 s`, what torque was applied to the object?

Answer 1

88N

Explanation:

first we need to find out the change in instantaneous velocity .
which is from 1 Hz to 6 Hz using formula
`omega`=`(2 piR)/T=vR`
`=>v=2pi/T`
substituting values of 1 Hz as `1/nu=T`
we get the values for instantanious velocity for 2 cases of 1 Hz and
6 Hz as 6.28`ms^(-1)` and `39.26ms^(-1)`
we use the Newton's law of motion according to which
`v=u+a*t`
substituting
u=6.28
v=39.26
and T=3
We get the value for 'a' Acceleration as 10.99`~=`11`ms^(-1)`
For torque =`m*a*R`
we substitute desired values and get 88 N