A wheel of radius 1-meter rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
Correct answer: C. √(π2 + 4)
- A. 2π
- B. √2 π
- C. √(π2 + 4)
- D. π
Explanation
Let's say that point A on the wheel is in contact with the ground at Point B initially. The wheel then completed one rotation and point A will again be in contact with the ground after 1 complete rotation. The distance traveled by the wheel will be equal to the circumference = 2πr.Now as per the question the ball moves half a revolution so the distance traveled will also be half of the circumference = 2πr/2 = πrSince after one complete revolution point A was in contact with the ground(lowest point on the wheel) after half a rotation point A will be halfway as well at the highest point on the wheel opposite to the point in contact with the ground. So, the distance between the ground and point A will be equal to the diameter = 2rNow we can form a right-angle triangle and use Pythagoras' theorem to calculate the displacement, which is the distance from the initial to the final position (refer to the diagram below)X2 = (πr)2 + (2r)2X2 = (π x 1)2 + (2 x 1)2X2 = π + 4X = √(π + 4)
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