# Find the velocity of (a) a cylinder and (b) a ring at the bottom of an inclined plane that is...

## Question:

Find the velocity of (a) a cylinder and (b) a ring at the bottom of an inclined plane that is 1.55 m high. The cylinder and ring start from rest and roll down the plane. Hint: Carry out the calculation including the unknown parameters, and they may cancel out.

## Rolling Motion:

When the object is in translational motion with the rotational motion, then such motion is known as rolling motion. In the rolling motion, the object has two kinetic energy that is rotational and translational.

Given data:

• Height of the slope {eq}\rm (H)= 1.55\ m {/eq}

(a)

Now, the kinetic energy at the bottom would be

{eq}\rm K = \dfrac{1}{2}mv^{2} +\dfrac{1}{2} Iw^{2} \\ K = \dfrac{1}{2}mv^{2} + \dfrac{1}{2} \left(\dfrac{mr^{2}}{2}\right) \left(\dfrac{v}{r}\right)^{2} \\ K = \dfrac{1}{2}mv^{2} + \dfrac{1}{4}mv^{2} \\ K = \dfrac{3}{4}mv^{2} \\ {/eq}

Where

• I is the moment of inertia of cylinder

Now, applying the energy conservation

{eq}\begin{align} \rm mgH &= \rm \dfrac{3}{4}mv^{2} \\ 9.8 \times 1.55 &= \rm \dfrac{3}{4}v^{2} \\ \rm v &= \rm 4.5 \ m/s \\ \end{align} {/eq}

(b)

Kinetic energy of the ring at the bottom owould be

{eq}\rm K = \dfrac{1}{2}mv^{2} +\dfrac{1}{2} Iw^{2} \\ K = \dfrac{1}{2}mv^{2} + \dfrac{1}{2} \left(mr^{2}\right) \left(\dfrac{v}{r}\right)^{2} \\ K = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}mv^{2} \\ K = mv^{2} \\ {/eq}

Where

• I is the moment of inertia of the ring

Now, applying the energy conservation

{eq}\begin{align} \rm mgH &= \rm mv^{2} \\ 9.8 \times 1.55 &= \rm v^{2} \\ \rm v &= \rm 3.897 \ m/s \\ \end{align} {/eq}