Question 50: Centre of Mass and Rotational Motion

Question

A wheel is free to rotate about a horizontal axis through O. A force of $$200 \mathrm{~N}$$ is applied at a point $$\mathrm{P} 2 \mathrm{~cm}$$ from the center $$\mathrm{O}$$. OP makes an angle of $$55^{\circ}$$ with $$\mathrm{x}$$ axis and the force is in the plane of the wheel making an angle of $$25^{\circ}$$ with the horizontal axis. What is the torque?

Accepted Answer

To calculate the torque ($ 	au $) produced by a force on a wheel, the formula used is $ 	au = rF \sin(	heta) $, where:
$ r $ is the distance from the axis to the point of application of the force, in meters,
$ F $ is the magnitude of the force, in Newtons,
$ 	heta $ is the angle between the line of action of the force and the lever arm or radius.
In this case:
The distance $ r = 2 $ cm $ = 0.02 $ meters (since $1$ cm $= 0.01$ m),
The magnitude of the force $ F = 200 $ N,
The angle $ 	heta $ is the difference between the angle formed by OP with the horizontal (x-axis) and the angle the force makes with the horizontal, which is $ 55^{\circ} - 25^{\circ} = 30^{\circ} $.
Thus, substituting these values into the torque formula:
$ 	au = 0.02 	imes 200 	imes \sin(30^{\circ}) $
Since $ \sin(30^{\circ}) = 0.5 $,
$ 	au = 0.02 	imes 200 	imes 0.5 = 2 	ext{ N m} $
Therefore, the torque produced by the force on the wheel is $ 2 $ Newton-meters, which corresponds to Option C (2 N m).

Answer

4 N m

Answer

3.2 N m

Answer

2 N m

Answer

3.4 N m

Centre of Mass and Rotational Motion

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