# Have a hoop (15kg) with a radius of 1.5m. Angular speed is 5 rad/s. What is the total KE...

## Question:

Have a hoop (15kg) with a radius of 1.5m. Angular speed is 5 rad/s. What is the total KE possessed by the hoop? The hoop begins to climb an incline which is at a 15 degree angle. Measured along the incline, how far up the incline does it rool before stopping if the incline has a coefficient of friction of .15?

## Rolling motion:

The motion which combines both rotation, as well as the translation motion, is known as the Rolling motion. Therefore the kinetic energy of the rolling motion is given by the following expression which includes both translational and rotational kinetic energy.

{eq}\rm K = \dfrac{1}{2}Iw^{2} + \dfrac{1}{2}mv^{2} {/eq}

where

• I is the moment of inertia
• w is the angular velocity
• v is the linear velocity
• m is the mass of the hoop

Given data:

• Mass of the hoop {eq}\rm (m) = 15 \ kg {/eq}
• Radius of the hoop {eq}\rm (r) = 1.5 \ m {/eq}
• Angular speed of the hoop {eq}\rm (w) = 5 \ rad/s {/eq}
• Angle of the incline {eq}\rm (\theta) = 15^\circ {/eq}

The moment of inertia of the hoop would be

{eq}\rm I = mr^{2} \\ I = 15 \times 1.5^{2} \\ I = 33.75 \ kg.m^{2} {/eq}

The kinetic energy of the hoop would be

{eq}\rm K = \dfrac{1}{2}Iw^{2} + \dfrac{1}{2}mv^{2} \\ K = [0.5 \times 33.75 \times 5^{2} ] + [0.5 \times 15 \times (5 \times 1.5)^{2} ] \\ K = 843.75 \ J {/eq}

When the hoop moves upward along the incline, then the kinetic energy of the hoop will convert into the potential energy of the hoop.

Let us say the hoop has moved by a distance "d" along the incline, therefore the height raised would be

{eq}\rm h = d\sin\theta \\ h = d\sin15^\circ {/eq}

Applying energy conservation,

{eq}\begin{align} \rm mgh + (\mu \times mg \cos\theta \times d)&= K \\ \rm (15 \times 9.8 \times d \sin15^\circ ) + (0.15 \times 15 \times 9.8 \cos15^\circ \times d)&= 843.75 \\ \rm d &= \rm 14.2\ m \\ \end{align} {/eq} 