Question

Q1.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

Explanation:

Given:

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

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:

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

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

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

So, final answer is:

$\boxed{ \bf\: KE = \dfrac{mgl \sin( \theta) }{1 + \frac{ {r}^{2} }{ {k}^{2} } } } </p><p>$

Answer 2

.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 mgl sin 0