JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 22)
Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $${{\sqrt x } \over 2}$$. Then, the value of x is _____________.
Answer
3
Explanation
I in both cases is about point of contact
Ring
mgh = $${1 \over 2}I{\omega ^2}$$
mgh = $$ = {1 \over 2}(2m{R^2}){{v_R^2} \over {{R^2}}}$$
$${v_R} = \sqrt {gh} $$
Solid cylinder
mgh = $${1 \over 2}I{\omega ^2}$$
mgh $$ = {1 \over 2}\left( {{3 \over 2}m{R^2}} \right){{v_C^2} \over {{R^2}}}$$
$${v_C} = \sqrt {{{4gh} \over 3}} $$
$${{{v_R}} \over {{v_C}}} = {{\sqrt 3 } \over 2}$$
Ring
mgh = $${1 \over 2}I{\omega ^2}$$
mgh = $$ = {1 \over 2}(2m{R^2}){{v_R^2} \over {{R^2}}}$$
$${v_R} = \sqrt {gh} $$
Solid cylinder
mgh = $${1 \over 2}I{\omega ^2}$$
mgh $$ = {1 \over 2}\left( {{3 \over 2}m{R^2}} \right){{v_C^2} \over {{R^2}}}$$
$${v_C} = \sqrt {{{4gh} \over 3}} $$
$${{{v_R}} \over {{v_C}}} = {{\sqrt 3 } \over 2}$$
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