# 2kg disc of radius 1m moment of inertia 0.5kgm^2 is released from top of 2m long incline having inclination 30degree. Find it’s velocity just before reaching the ground .??

Jul 18, 2018

Given

Mass of the disc $M = 2 k g$

Radius of the disc $r = 1 m$

Length of incline $l = 2 m$

Moment inertia of the disc $I = 0.5 k g {m}^{2}$

Angle of inclination of the incline $\theta = {30}^{\circ}$
When the disc reaches at the ground let the linear velocity of the disc be $v$ and angular velocity be $\omega$. At the ground whole of it's gravitational PE will be transfromed into rotational and translational KE.

So

$\frac{1}{2} I {\omega}^{2} + \frac{1}{2} M {v}^{2} = M g l \sin \theta$

$\implies \frac{1}{2} I {v}^{2} / {r}^{2} + \frac{1}{2} M {v}^{2} = M g l \sin \theta$

$\implies I {v}^{2} / {r}^{2} + M {v}^{2} = 2 M g l \sin \theta$

$\implies {v}^{2} \left(\frac{I}{r} ^ 2 + M\right) = 2 M g l \sin \theta$

$\implies {v}^{2} \left(\frac{0.5}{1} ^ 2 + 2\right) = 2 \times 2 \times 9.8 \times 2 \times \sin {30}^{\circ}$

$\implies {v}^{2} \times \frac{5}{2} = 2 \times 2 \times 9.8 \times 2 \times \frac{1}{2}$

$\implies {v}^{2} = 2 \times 2 \times 9.8 \times 2 \times \frac{1}{2} \times \frac{2}{5}$

$= v = \sqrt{9.8 \times 1.6} = 2.8 \cdot \sqrt{2} m \text{/} s$

Jul 18, 2018

The speed is $= 3.61 m {s}^{-} 1$

#### Explanation:

Another approach is to calculate the acceleration of the disc down the incline.

Let the coefficient if friction detween the disc and the incline be $= \mu$

Then,

The acceleration down the plane is

$m g \sin \theta - \mu m g \cos \theta = m a$

$g \sin \theta - \mu g \cos \theta = a$......................$\left(1\right)$

Taking moments about the center of the disc

$\tau = I \alpha$

$\alpha = \frac{a}{r}$

$r \mu m g \cos \theta = \frac{1}{2} m {r}^{2} \cdot \frac{a}{r}$

$\mu = \frac{a}{2 g \cos \theta}$............................$\left(2\right)$

Plugging this value in equation $\left(1\right)$

$g \sin \theta - \frac{a}{2 g \cos \theta} g \cos \theta = a$#

Therefore,

$\frac{3}{2} a = g \sin \theta$

The acceleration is $a = \frac{2}{3} g \sin \theta$

The distance is $s = 2 m$

Apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

${v}^{2} = 0 + 2 a s$

$v = \sqrt{2 \cdot \frac{2}{3} g \sin \theta \cdot s}$

$v = \sqrt{\frac{4}{3} \cdot 9.8 \cdot \sin 30 \cdot 2} = \sqrt{4 \cdot \frac{9.8}{3}} = 3.61 m {s}^{-} 1$

P.S There is a problem, the moment of inertia of the disc about the center is $I = \frac{1}{2} m {r}^{2} = \frac{1}{2} \cdot 2 \cdot {1}^{2} = 1 k g {m}^{2}$ and not $0.5 k g {m}^{2}$