A uniform disc of radius R mass m , rolls over a horizontal surface as shown. At given instant t = t 0 the velocities of two points P , Q on the diameter POQ are given as shown in the figure. Given

Question

A uniform disc of radius R mass m , rolls over a horizontal surface as shown. At given instant t = t 0 ​ the velocities of two points P , Q on the diameter POQ are given as shown in the figure. Given that θ = tan − 1 ( 2 1 ​ )

TestHub · Accepted Answer

Combining the velocity vectors of the diametrically opposite points $P$ and $Q$ given in the figure: $v_{C} = V cos θ$ and $ω R = \frac{V c o s θ}{2}$ .

Since $v_{C} > ω R$ , the contact point slides forward - forward slipping, (B).

$K E = \frac{1}{2} m v_{C}^{2} + \frac{1}{2} \cdot \frac{m R ^{2}}{2} ω^{2} = \frac{m V ^{2} c o s ^{2} θ}{2} + \frac{m V ^{2} c o s ^{2} θ}{16} = \frac{9}{16} m V^{2} cos^{2} θ$ - (C).

Highest point $S$ : $v_{S} = v_{C} + ω R = \frac{3 V c o s θ}{2}$ - (D).

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