Question

A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tangentially at the rim (a) Determine the rotational kinetic energy oh the hoop (b) What is instantaneous change rate of the kinetic energy?

Answer 1

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Step-by-step explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

`K.E =(1)/(2)*I\omega^2`

`K.E=(1)/(2)* mr^2*\omega^2`

`K.E=(1)/(2)*2*(0.5)^2*(3)^2`

`K.E=2.25\ J`

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

`K.E=(1)/(2)mv^2`

On differentiating

`(K.E)/(dt)=(1)/(2)m*2v*(dv)/(dt)`

`(K.T)/(dt)=mva`....(I)

Using newton's second law

`F = ma`

`a= (F)/(m)`

`a=(10)/(2)`

`a=5 m/s^2`

Put the value of a in equation (I)

`(K.E)/(dt)=mva`

`(K.E)/(dt)=mr\omega a`

`(K.E)/(dt)=2*0.5*3*5`

`(K.E)/(dt)=15\ J/s`

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.