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A child stands at the centre of a turn table with his two arms outstretched. The turn table is set rotating with an angular speed of $40 \mathrm{rpm}$. How much is the angular speed of the child, if he folds his hands back reducing the M.I. to $2 / 5$ time the initial value? Assume that the table rotates without friction (b) Show that the child's new K.E. of rotation is more than the initial K.E. of rotation. How do you account for this increase in K.E.?
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Here, $\omega_1=40 \mathrm{rev} / \mathrm{min}, I_2=\frac{2}{5} I_1$
$\because \quad$ No external torque is acting on the system
$\because L=I \omega=$ constant
$\therefore \quad I_1 \omega_1=I_2 \omega_2$
$\omega_2=\frac{I_1}{I_2} \omega_1=\frac{5}{2} \times 40=100 \mathrm{rpm} ;$
$\frac{\text { FinalK.E. }}{\text { InitialK.E. }}=\frac{\frac{1}{2} I_2 \omega_2^2}{\frac{1}{2} I_1 \omega_1^2}=\left(\frac{I_2}{I_1}\right)\left(\frac{\omega_2}{\omega_1}\right)^2$
$$
=\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}=2.5
$$
$\therefore \quad$ Final K.E. $=2.5$ (Initial K.E.)
This increase in K.E is due to the fact that child spends his internal energy in folding his hands which is converted into K.E.
$\because \quad$ No external torque is acting on the system
$\because L=I \omega=$ constant
$\therefore \quad I_1 \omega_1=I_2 \omega_2$
$\omega_2=\frac{I_1}{I_2} \omega_1=\frac{5}{2} \times 40=100 \mathrm{rpm} ;$
$\frac{\text { FinalK.E. }}{\text { InitialK.E. }}=\frac{\frac{1}{2} I_2 \omega_2^2}{\frac{1}{2} I_1 \omega_1^2}=\left(\frac{I_2}{I_1}\right)\left(\frac{\omega_2}{\omega_1}\right)^2$
$$
=\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}=2.5
$$
$\therefore \quad$ Final K.E. $=2.5$ (Initial K.E.)
This increase in K.E is due to the fact that child spends his internal energy in folding his hands which is converted into K.E.
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