Question 748: Centre of Mass and Rotational Motion

Question

A ring and disc are initially at rest, side by side, at the top of an  inclined plane which makes an angle $${60^ \circ }$$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $$\left( {2 - \sqrt 3 } \right)/\sqrt {10} \,\,s,$$ then the height of the top of the inclined plane, in metres is ______________ . Take $$g = 10\,\,m{s^{ - 2}}.$$

Accepted Answer

Given, h = height of the top of inclined plane = ?, $$	heta$$ = 60$$^\circ$$, g = 10 m s
$$-$$2
and time difference between ring and disc reaching ground = $${{2 - \sqrt 3 } \over {\sqrt {10} }}$$ s
We know that, $$a = {{g\sin 	heta } \over {1 + {I \over {M{R^2}}}}}$$
For ring, $$I = M{R^2}$$ and for disc $$I = {3 \over 2}M{R^2}$$.
So, $${a_{ring}} = {{g\sin 	heta } \over 2}$$ and $${a_{disk}} = {{2g\sin 	heta } \over 3}$$
Now using $$s = {1 \over 2}a{t^2}$$
$${s_{ring}} = {1 \over 2}\left( {{{g\sin 	heta } \over 2}} ight)t_1^2 \Rightarrow {h \over {\sin 	heta }} = {1 \over 2}\left( {{{g\sin 	heta } \over 2}} ight)t_1^2$$ ...... (1)
$${s_{disk}} = {1 \over 2}\left( {{{2g\sin 	heta } \over 3}} ight)t_2^2 \Rightarrow {h \over {\sin 	heta }} = {1 \over 2}\left( {{{2g\sin 	heta } \over 3}} ight)t_2^2$$ ...... (2)
Given, $${t_1} - {t_2} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$$
From Eq. (1), we have $${t_1} = \sqrt {{{4h} \over {g{{\sin }^2}	heta }}} $$
and from Eq. (2), we have $${t_2} = \sqrt {{{3h} \over {g{{\sin }^2}	heta }}} $$
So, $$\sqrt {{{4h} \over {g{{\sin }^2}	heta }}}  - \sqrt {{{3h} \over {g{{\sin }^2}	heta }}}  = {{2 - \sqrt 3 } \over {\sqrt {10} }}$$
$$ \Rightarrow \sqrt {{{4h} \over {10{{\sin }^2}60^\circ }}}  - \sqrt {{{3h} \over {10{{\sin }^2}60^\circ }}}  = {{2 - \sqrt 3 } \over {\sqrt {10} }}$$
$$ \Rightarrow \sqrt {{{16h} \over {3 	imes 10}}}  - \sqrt {{{4h} \over {10}}}  = {{2 - \sqrt 3 } \over {\sqrt {10} }}$$
$$ \Rightarrow {{4\sqrt h } \over {\sqrt 3 }} - 2\sqrt h  = 2 - \sqrt 3 $$
$$ \Rightarrow \sqrt h \left( {{4 \over {\sqrt 3 }} - 2} ight) = (2 - \sqrt 3 )$$
$$ \Rightarrow \sqrt h {{(4 - 2\sqrt 3 )} \over {\sqrt 3 }} = (2 - \sqrt 3 )$$
$$ \Rightarrow \sqrt h  	imes {2 \over {\sqrt 3 }}(2 - \sqrt 3 ) = (2 - \sqrt 3 ) \Rightarrow \sqrt h  = {{\sqrt 3 } \over 2}$$
$$ \Rightarrow h = {3 \over 4} = 0.75$$ m

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type