A) \[\frac{3{{v}^{2}}}{5R}\]
B) \[\frac{2{{v}^{2}}}{3R}\]
C) \[\frac{5{{v}^{2}}}{6R}\]
D) \[\frac{{{v}^{2}}}{R}\]
Correct Answer: C
Solution :
[c] The centre of the wheel is moving with constant speed on a circular path of radius\[6R\]. Hence it has a centripetal acceleration of \[{{a}_{c}}=\frac{{{v}^{2}}}{6R}\]directed towards the centre of curvature of the convex surface. With respect to the centre of the wheel the contact point has acceleration equal to \[\frac{{{v}^{2}}}{R}\] directed towards the centre of the wheel. \[\therefore \] Acceleration of the contact point in reference frame of ground is \[{{a}_{P}}=\frac{{{v}^{2}}}{R}-\frac{{{v}^{2}}}{6R}=\frac{5{{v}^{2}}}{6R}\]
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