# A 1-kg mass hangs by a string from a disk with radius 4.6 cm which has a rotational inertia of 5...

## Question:

A 1-kg mass hangs by a string from a disk with radius 4.6 cm which has a rotational inertia of 5 x {eq}10^{-5} kg.m^2 {/eq}. After it falls a distance of 0.8 meters, how fast is it going to the nearest hundredth of a m/s?

## Rotational motion;

The object in the rotation having the rotational kinetic energy and it is directly proportional to the moment of inertia and square of the angular velocity of the motion.

Given Data:

• Mass, {eq}\rm (m) = 1 \ kg {/eq}
• Moment of inertia of disk {eq}\rm (I) = 5 \times 10^{-5} kg-m^{2} {/eq}
• Radius of the disk {eq}\rm (r) = 4.6 \ cm \\ {/eq}
• Height of the fall {eq}\rm (h) = 0.8 \ m {/eq}

Now, appplying the conservation of energy

{eq}\begin{align} \rm mgh &=\rm \dfrac{1}{2}mv^{2} + \dfrac{1}{2}I \left(\dfrac{v}{r} \right)^{2} \\ 1 \times 9.8 \times 0.8 &= \rm [ 0.5 \times 1v^{2} ] + [ 0.5 \times (5 \times 10^{-5}) \times \left(\dfrac{v}{0.046 \ m } \right)^{2} ] \\ \rm v &= \rm 3.91 \ m/s \\ \end{align} {/eq} 