Question

A thin uniform disc of mass m and radius r is released from a rough inclined plane of inclination theta .friction is sufficient for pure rolling .rotational kinetic energy of disc when it falls down a distance l over the plane is

Answer 1

Given:

A thin uniform disc of mass m and radius r is released from a rough inclined plane of inclination $\theta$.

To find:

Rotational kinetic energy?

Calculation:

Let rotational kinetic energy be KE :

$\rm \: KE = \dfrac{1}{2} I { \omega}^{2}$

$\rm \implies\: KE = \dfrac{1}{2} \times (m {k}^{2}) \times { \omega}^{2}$

• "k" is Radius of Gyration.

$\rm \implies\: KE = \dfrac{1}{2} \times (m {k}^{2}) \times {( \dfrac{v}{r} )}^{2}$

$\rm \implies\: KE = \dfrac{1}{2} m {v}^{2} \bigg( \dfrac{ {k}^{2} }{ {r}^{2} } \bigg )$

For inclined plane rolling:

$\rm \implies\: KE = \dfrac{1}{2} m \times \dfrac{2gh}{1 + \frac{ {k}^{2} }{ {r}^{2} } } \times \bigg( \dfrac{ {k}^{2} }{ {r}^{2} } \bigg )$

$\rm \implies\: KE = \dfrac{1}{2} m \times \dfrac{2gl \sin( \theta) }{1 + \frac{ {k}^{2} }{ {r}^{2} } } \times \bigg( \dfrac{ {k}^{2} }{ {r}^{2} } \bigg )$

$\rm \implies\: KE = \dfrac{mgl \sin( \theta) }{1 + \frac{ {r}^{2} }{ {k}^{2} } }$

• For a disc, value of k²/r² = ½.

$\rm \implies\: KE = \dfrac{mgl \sin( \theta) }{3}$

So, final answer is:

$\boxed{ \bf\: KE = \dfrac{mgl \sin( \theta) }{3} }$

Answer 2

☆Concept:

》Whenever any object is released from height its potential energy will get change to some other form of energy.

》As it is the case of pure rolling there will be no energy loss takes place as friction during moving down.

》So all the potential energy will converted to kinetic energy of disc at bottom.

》And it will be consist of rotational KE + translational KE.

☆Solution:

see the attached image during solving!

Given: disc;

• mass - m
• radius- r
• distance of falling- l

》$\rm Inertia_{disc} = \dfrac{1}{2}mr^{2}$

By applying energy conservation:

• $\rm mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega ^2$

• $\rm mgh = \dfrac{1}{2}m\omega ^2 r^2 + \dfrac{1}{2}(\dfrac{1}{2}mr^{2})\omega ^2$...(pure rolling $\ v=\omega r$)

• $\rm glsin\theta = \dfrac{\omega ^2 r^2}{2} + \dfrac{\omega ^2r^2}{4}$

• $\rm glsin\theta = \dfrac{3\omega ^2 r^2}{4}$

• $\bf \omega ^2 = \dfrac{4glsin\theta}{3r^2}$

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Rotational KE(disc) = $\bf \dfrac{1}{2}I\omega ^2$

• $\rm \dfrac{1}{2}( \dfrac{1}{2}mr^2) \omega ^2$

• $\rm \dfrac{1}{4}mr^2( \dfrac{4glsin\theta}{3r^2})$

• $\bf \dfrac{mglsin\theta}{3}$

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