Question #b1c2e

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Question #b1c2e

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Answer 1 · 2017-01-15T01:39:54

(b)

Explanation:

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

Potential energy $P E = m g h$ $PE=mgh$
Rotational Kinetic Energy given by $K E_{R} = \frac{1}{2} I ω^{2}$ $KE_R=1/2Iomega^2$
where moment of inertia $I = \frac{1}{2} m R^{2}$ $I=1/2mR^2$ and angular velocity $ω = \frac{v}{R}$ $omega=v/R$
Translational Kinetic energy $K E_{T} = \frac{1}{2} m v^{2}$ $KE_T=1/2mv^2$

(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 $P E$ $PE$
(b) At the bottom of hill all the $P E$ $PE$ gets converted into sum of $K E_{T}$ $KE_T$ and $K E_{R}$ $KE_R$

By Law of Conservation of energy

$T E_{(a)} = T E_{(b)}$ $TE_"(a)"=TE_"(b)"$

$m g h = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$ $mgh=1/2Iomega^2+1/2mv^2$
$\Rightarrow m g h = \frac{1}{2} (\frac{1}{2} m R^{2}) {(\frac{v}{R})}^{2} + \frac{1}{2} m v^{2}$ $=>mgh=1/2(1/2mR^2)(v/R)^2+1/2mv^2$
$\Rightarrow g h = \frac{1}{4} v^{2} + \frac{1}{2} v^{2}$ $=>gh=1/4v^2+1/2v^2$
$\Rightarrow g h = \frac{3}{4} v^{2}$ $=>gh=3/4v^2$
$\Rightarrow v = \sqrt{\frac{4}{3} g h}$ $=>v=sqrt(4/3gh)$

Inserting given values we get
$v = \sqrt{\frac{4}{3} \times 9.81 \times 10}$ $v=sqrt(4/3xx 9.81xx10)$
$v \approx 11.43 m s^{- 1}$ $vapprox11.43ms^-1$

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Question #b1c2e

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