Question

Compute the rotational KE of a 25 kg wheel rotating at 6.0 revs/s if the radius of gyration of the wheel is 22cm.​

Answer 1

To find:

The rotational KE of a 25 kg wheel rotating at 6.0 revs/s if the radius of gyration of the wheel is 22cm. ?

Calculation:

The general expression for Rotational Kinetic Energy of a body is:

$\boxed{ \bold{\therefore \: KE = \dfrac{1}{2} I { \omega}^{2} }}$

• "I" refers to Moment Of Inertia , $\omega$ represents the angular velocity.

• Moment Of Inertia can also be represented as product of mass and square of radius of gyration.

$\boxed{ \bold{\implies\: KE = \dfrac{1}{2} (m {k}^{2}) { \omega}^{2} }}$

Putting available values in SI units:

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{25 \times { (\dfrac{22}{100} )}^{2} \bigg \}{ (6 \times 2\pi )}^{2}$

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{25 \times (\dfrac{484}{10000} ) \bigg \}{ (6 \times 2\pi )}^{2}$

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{ \dfrac{484}{400} \bigg \}{ (6 \times 2\pi )}^{2}$

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{ \dfrac{121}{100} \bigg \}{ (6 \times 2\pi )}^{2}$

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{ 1.21 \bigg \}{ (6 \times 2\pi )}^{2}$

$\sf\implies\: KE = \dfrac{1}{2} \bigg \{ 1.21 \bigg \} \times 36 \times 4 {\pi}^{2}$

$\sf\implies\: KE = \bigg \{ 1.21 \bigg \} \times 18 \times 4 {\pi}^{2}$

$\sf\implies\: KE = 859.83 \: joule$

$\sf\implies\: KE \approx 860\: joule$

So, Rotational Kinetic Energy is approximately 860 Joules.