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
B
0.23
C
0.05
D
0.87

Solution

The correct option is D 0.87Rotation 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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