JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 4)
A person of 80 kg mass is standing on the rim
of a circular platform of mass 200 kg rotating
about its axis at 5 revolutions per minute (rpm).
The person now starts moving towards the
centre of the platform. What will be the
rotational speed (in rpm) of the platform when
the person reaches its centre _________.
Answer
9
Explanation
$${I_1}{\omega _1} = {I_2}{\omega _2}$$
$$ \Rightarrow $$ $$\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$$
$$ \Rightarrow $$ $$\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$$
$$ \Rightarrow $$ $$\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$$
$$ \Rightarrow $$ $${\omega _2} = \left( {1.8} \right){\omega _1}$$
$$ \Rightarrow $$ 2$$\pi $$f2 = 2$$\pi $$f1 $$ \times $$ 1.8
$$ \Rightarrow $$ f2 = 5 $$ \times $$ 1.8 = 9
$$ \Rightarrow $$ $$\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$$
$$ \Rightarrow $$ $$\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$$
$$ \Rightarrow $$ $$\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$$
$$ \Rightarrow $$ $${\omega _2} = \left( {1.8} \right){\omega _1}$$
$$ \Rightarrow $$ 2$$\pi $$f2 = 2$$\pi $$f1 $$ \times $$ 1.8
$$ \Rightarrow $$ f2 = 5 $$ \times $$ 1.8 = 9
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