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Answer 1

Answer on Question#56567 - Physics - Mechanics - Relativity

5-7 A wheel of radius 20 cm is pushed to move-it on a rough horizontal surface. It is found to move through a distance of 60 cm on the road during the time it completes one revolution about the centre. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is :

(A) Along the velocity of the wheel

(B) Opposite to the velocity of the wheel

(C) Perpendicular to the velocity of the wheel

(D) Zero

Rigid Bodies and Rotational Motion

Solution:

The circumference of the wheel is

$$L = 2 \pi \cdot 20 \mathrm{cm} \approx 125.66 \mathrm{cm}$$

It is greater than the distance it moved through on the road (60cm). It means that the wheel slips. Therefore, at the point of touch the velocity of the point on the wheel is opposite to the velocity of the wheel's centre and hence the frictional force (which is opposite to the velocity of this point) is along the velocity of the wheel.

Answer: (A).

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