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 =** KE _{L} = (1/2) m v^{2} **

Let’s Convert that equation to its angular analog:

KE = (1/2) (mr^{2}).(v^{2}/r^{2}) = (1/2) (mr^{2}).(v/r)^{2} = (1/2) I ω^{2} ** KE _{R} = (1/2) I ω^{2} **

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 r^{2}

**KE _{R} = (1/2) I ω^{2} ** = (1/2) (2/5) m r

^{2}.

**ω**= (1/2) (2/5) 100 2

^{2}^{2}.

**10**= 800 J

^{2}**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 r^{2}

**KE _{R} = (1/2) I ω^{2} ** = (1/2) m r

^{2}.

**ω**= (1/2) 10 (0.5)

^{2}^{2}.

**(**)

**80π****= 78.8 x 10**

^{2}^{3}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?**