JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 22)
Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.
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Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s$$-$$1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s$$-$$1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]
Answer
20
Explanation
$$\alpha = {\tau \over I} = {{F.R.} \over {m{R^2}/2}} = {{2F} \over {mR}}$$
$$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$$ rad/s2
$${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $$
$${(50)^2} = {0^2} + 2(10)\Delta \theta $$
$$ \Rightarrow \Delta \theta = {{2500} \over {20}}$$
$$\Delta \theta = 125$$ rad
No. of revolution $$ = {{125} \over {2\pi }} \approx 20$$ revolution
$$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$$ rad/s2
$${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $$
$${(50)^2} = {0^2} + 2(10)\Delta \theta $$
$$ \Rightarrow \Delta \theta = {{2500} \over {20}}$$
$$\Delta \theta = 125$$ rad
No. of revolution $$ = {{125} \over {2\pi }} \approx 20$$ revolution
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