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Question

A wheel of mass $$10kg$$ and radius $$20cm$$ is rotating at an angular speed pf $$100$$ revolution$$/$$min when the motor is turned off. neglecting the friction at the axle, Calculate the force that must be applied tangentially to the wheel to bring it to rest in $$10$$ revolutions$$?$$


A
0.02
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B
0.23
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C
0.05
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D
0.87
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Solution

The correct option is D 0.87
Rotation of the wheel $$= 100$$ $$revolution / min$$ or $$= 5/3$$
 $$revolution/second$$ $$.$$
Now$$ ,w' = 0$$ 
Total rev. $$10$$revolution 
Now we have to find $$\alpha=? $$
We know the formula$$,$$
$$w' = {w^2} - 2\alpha \theta $$
$$0 = (10/6)^2 - 2 × α × 10$$
$$0 = 100/36 - 20α$$
$$α = 5/36 rev/second^2$$
or$$, α = 2 π × (5/36)$$
$$α = 10π / 36 radian/second^2 .$$
Now$$,$$ we have to moment of inertia of the wheel$$.$$
$$I = \frac{{m{r^2}}}{2}$$
$$I = 1/2 (0.2^2 × 10)$$
$$I = 0.2 kg.m^2$$
Since the force is F, Now torque $$,$$
$$F = F.r$$
$$0.2 F$$
Now torque $$= I α$$
$0.20 / (10 π /36)44
$$= 2 π /36$$
$$0.2 F = 2 π /36$$
$$F = 2 π/7.2$$
$$F = 0.87 N $$
Hence,
option $$(D)$$ is correct answer.

Physics

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