High School Physics + more

# Numerical problem on Rotational Kinetic Energy

Last updated on July 12th, 2023 at 03:56 pm

In this post, we will review the formula of Rotational Kinetic Energy and then solve Numerical problems based on Rotational Kinetic Energy.

## Rotational Kinetic Energy formula from the linear kinetic energy formula

The equation for linear kinetic energy = KEL = (1/2) m v2

Let’s Convert that equation to its angular analog:
KE = (1/2) (mr2).(v2/r2) = (1/2) (mr2).(v/r)2 = (1/2) I ω2
KER = (1/2) I ω2

The equation of linear kinetic energy can be converted to the angular analog equation when the motion is rotational. The angular velocity ω takes the place of the linear velocity v, and the momentum of inertia (I) takes the place of the mass m.

You may wish to read a separate post specifically on the Detailed step-by-step derivation of the Rotational KE formula for class 11.

## Numerical problem based on Rotational Kinetic Energy

While solving the Numericals you can have a quick look at this post: Moment of Inertia of different bodies

1 ) A 100-kg solid sphere with a radius equal to 2.0 m is rotating at 10.0 radians/s, what is its rotational kinetic energy?

Solution:

r= 2 m

m = 100 kg

For a solid sphere, the momentum of inertia I = (2/5) m r2

KER = (1/2) I ω2 = (1/2) (2/5) m r2 . ω2 = (1/2) (2/5) 100 22 . 102 = 800 J

2 ) How much rotational kinetic energy does a spinning tire of mass 10.0 kg and radius 0.50 m have if it’s spinning at 40.0 rotations/s?

Solution:

ω = 40 rps = 40 x 2π rad/sec = 80π rad/sec

r= 0.5 m

m = 10 kg

For a spinning tire, the momentum of inertia I = m r2

KER = (1/2) I ω2 = (1/2) m r2 . ω2 = (1/2) 10 (0.5)2 . (80π)2 = 78.8 x 103 J

To revise the formulas you can have a quick look at this post: Moment of Inertia of different bodies