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

Answer:

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

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

$K.E =\dfrac{1}{2}\timesI\omega^2$

$K.E=\dfrac{1}{2}\times mr^2\times\omega^2$

$K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2$

$K.E=2.25\ J$

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

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

On differentiating

$\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}$

$\dfrac{K.T}{dt}=mva$....(I)

Using newton's second law

$F = ma$

$a= \dfrac{F}{m}$

$a=\dfrac{10}{2}$

$a=5 m/s^2$

Put the value of a in equation (I)

$\dfrac{K.E}{dt}=mva$

$\dfrac{K.E}{dt}=mr\omega a$

$\dfrac{K.E}{dt}=2\times0.5\times3\times5$

$\dfrac{K.E}{dt}=15\ J/s$

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