JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 18)
A cubical block of side 30 cm is moving with velocity 2 ms−1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular
velocity (in rad/s) of the block immediately after it hits the bump, is :
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5.0
6.7
9.4
13.3
Explanation
Before hitting point 0,
angular moment = mv $$ \times $$ $${a \over 2}$$
After hitting point 0,
Angular momentum = $${\rm I}\omega $$
$$ \therefore $$ $${\rm I}\omega $$ = $${{mva} \over 2}$$
$$ \Rightarrow $$ $$\omega $$ = $${{mva} \over {2{\rm I}}}$$
$${\rm I}$$ = moment of inertia about edge,
= $${{m{a^2}} \over 6} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}$$
= $${{m{a^2}} \over 6} + {{m{a^2}} \over 2}$$
= $${{2m{a^2}} \over 3}$$
$$ \therefore $$ $$\omega $$ = $${{mva} \over {2 \times {{2m{a^2}} \over 3}}}$$ = $${{3v} \over {4a}}$$ = $${{3 \times 2} \over {4 \times 0.3}}$$ = 5 rad/s
angular moment = mv $$ \times $$ $${a \over 2}$$
After hitting point 0,
Angular momentum = $${\rm I}\omega $$
$$ \therefore $$ $${\rm I}\omega $$ = $${{mva} \over 2}$$
$$ \Rightarrow $$ $$\omega $$ = $${{mva} \over {2{\rm I}}}$$
$${\rm I}$$ = moment of inertia about edge,
= $${{m{a^2}} \over 6} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}$$
= $${{m{a^2}} \over 6} + {{m{a^2}} \over 2}$$
= $${{2m{a^2}} \over 3}$$
$$ \therefore $$ $$\omega $$ = $${{mva} \over {2 \times {{2m{a^2}} \over 3}}}$$ = $${{3v} \over {4a}}$$ = $${{3 \times 2} \over {4 \times 0.3}}$$ = 5 rad/s
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