Question

A wheel of perimeter4 pie metres is rolling on a horizontal surface . the displacement of the point of contact of the wheel and ground when the wheel completes one quarter of revolution is
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Answer 1

Answer:

a wheel with a perimeter (circumference) of 4 pi, so the diameter is 4, (pi x dia = 4 pi) and radius 2

when the wheel travels one quarter of revolution, the point of contact is 1/4 x 4pi = pi chord distance from the ground and let the straight line distance be A

draw an imaginary triangle by three lines resp. the diamenter of the circle, side A and a line joining the remaining two points of the diameter and this line A, which is also of the same length A

by calculation, after applying h2 = p2 + b2 (pythagoras theorum)

A2 + A2 = diameter2

2A2 = 4x4 = 16

so, A2 = 8

hence A = 2.828

hence 2.828 is the linear distance from the original point of contact when the wheel turns a quarter revolution..

Answer 2

Answer:

answer is as simple as 2√2 m

hope it helps...

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