# Question #b1c2e

Jan 15, 2017

(b)

#### Explanation:

When disc of mass $m$ and radius $R$ rolls down a hill with a velocity $v$ of its centre of mass, without slipping, it has three types of energies associated with it at any instant of time

1. Potential energy $P E = m g h$
2. Rotational Kinetic Energy given by $K {E}_{R} = \frac{1}{2} I {\omega}^{2}$
where moment of inertia $I = \frac{1}{2} m {R}^{2}$ and angular velocity $\omega = \frac{v}{R}$
3. Translational Kinetic energy $K {E}_{T} = \frac{1}{2} m {v}^{2}$

(We have ignored friction in this statement.)

We also know that

(a) Initial energy at the top of hill when disc is at rest it has only $P E$
(b) At the bottom of hill all the $P E$ gets converted into sum of $K {E}_{T}$ and $K {E}_{R}$

By Law of Conservation of energy

$T {E}_{\text{(a)"=TE_"(b)}}$

$m g h = \frac{1}{2} I {\omega}^{2} + \frac{1}{2} m {v}^{2}$
$\implies m g h = \frac{1}{2} \left(\frac{1}{2} m {R}^{2}\right) {\left(\frac{v}{R}\right)}^{2} + \frac{1}{2} m {v}^{2}$
$\implies g h = \frac{1}{4} {v}^{2} + \frac{1}{2} {v}^{2}$
$\implies g h = \frac{3}{4} {v}^{2}$
$\implies v = \sqrt{\frac{4}{3} g h}$

Inserting given values we get
$v = \sqrt{\frac{4}{3} \times 9.81 \times 10}$
$v \approx 11.43 m {s}^{-} 1$