Question 272: Centre of Mass and Rotational Motion

Question

A rectangular solid box of length 0.3 m is held
horizontally, with one of its sides on the edge
of a platform of height 5m. When released, it
slips off the table in a very short time t = 0.01s,
remaining essentially horizontal. The angle by
which it would rotate when it hits the ground
will be (in radians) close to :-

Accepted Answer

Angular impulse = change in angular momentum.
$$mg{l \over 2} 	imes 0.01 = {{m{l^2}} \over 3}\omega $$
$$\omega  = {{3g 	imes 0.01} \over {2l}} = {{3 	imes 10 	imes 0.01} \over {2 	imes 0.3}}$$
$$ = $$ $${1 \over 2} = 0.5\,rad/s$$
Time taken by rod to hit the ground
$$t = \sqrt {{{2h} \over g}}  = \sqrt {{{2 	imes 5} \over {10}}}  = 1\,\sec .$$
In this time angle rotate by rod
$$	heta $$ = $$\omega $$t = 0.5 ÃÂ 1 = 0.5 radian

Answer

0.28

Answer

0.02

Answer

0.3

Answer

0.5

Centre of Mass and Rotational Motion

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