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BITSAT 2013 Paper

Section: Physics

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Question : 12 of 150

Marks: +1, -0

A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc –

velocity is v, acceleration is zero
velocity is zero, acceleration is zero
velocity is v, acceleration is $v^2$ /R
velocity is zero, acceleration is $v^2$ /R

Solution:

As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure

v_{net}

(for lowest point)

= v - Rω = v - v = 0 and acceleration =

\frac{v^2}{R} + 0

=

\frac{v^2}{R}

(since linear speed is constant)

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