Rolling Energy Partition

Question

Total kinetic energy of rigid body rolling without slipping is:

RankGuru · Accepted Answer

For an object rolling without slipping, the total kinetic energy ($K_{total}$) is the sum of:
1. **Translational KE** of the center of mass: $K_{trans} = \frac{1}{2} m v^2$.
2. **Rotational KE** about the center of mass: $K_{rot} = \frac{1}{2} I \omega^2$.
$K_{total} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$
Using $v = r \omega$, we can also write $K_{total} = \frac{1}{2} m v^2 (1 + \frac{k^2}{r^2})$ where $k$ is the radius of gyration.

Rank Guru

Rolling Energy Partition

Select the correct option:

Solution

🔒 Solution Hidden from View

More rolling energy partition Practice Questions

A solid sphere (I = 2/5 MR²) rolls without slipπng at speed v. Total kinetic energy?

About This Question