Kinematics Question 122
Question: 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 it’s [BCECE 2005]
Options:
A) $ 2\pi $
B) $ \sqrt{2}\pi $
C) $ \sqrt{{{\pi }^{2}}+4} $
D) $ \pi $
Show Answer
Answer:
Correct Answer: C
Solution:
Horizontal distance covered by the wheel in half revolution = $ \pi R. $ So the displacement of the point which was initially in contact with ground = AA’ = $ \sqrt{{{(\pi R)}^{2}}+{{(2R)}^{2}}} $ = $ R\sqrt{{{\pi }^{2}}+4} $
$ =\sqrt{{{\pi }^{2}}+4} $
$ (AsR=1m) $
