Numerical problem on Rotational Kinetic Energy

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.

Numerical problem based on Rotational Kinetic Energy

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:

ω = 10 rad/sec

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 do 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

3 )

How much rotational kinetic energy do a spinning tire of mass 12 kg and radius 0.80 m have if it’s spinning at 200.0 radians/s?

4 )

How much work do you do to spin a tire, which has a mass of 5.0 kg and a radius of 0.40 m, from 0.0 radians/s to 100.0 radians/s?

5 )

How much work do you do to spin a hollow sphere, which has a mass of 10.0 kg and a radius of 0.50 m, from 0.0 radians/s to 200.0 radians/s?

Numerical problem on Rotational Kinetic Energy
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