First slide

Rolling motion

Question

A solid sphere is set into rotation at an angular velocity and it is then placed on a rough horizontal surface. The ratio of distances covered by rotational and translational motions up to the start of the pure rolling is (Assume uniformly accelerated motion up to start of pure rolling):

Difficult

A

5/2
B

7/2
C

7/4
D

9/2

Solution

Angular momentum about A is conserved.
$\begin{array}{l} L_{1} = \frac{2}{5} m R^{2} ω_{0} \\ L_{2} = \frac{2}{5} m R^{2} ω + m R^{2} ω = \frac{7}{5} m R^{2} ω \\ L_{1} = L_{2} \\ ω = \frac{2}{7} ω_{0} \\ ω_{0} > ω \end{array}$
$and v > v_{0}$
Sphere accelerates forward and rotation decelerates.
$\begin{array}{l} ω = ω_{0} - α t \\ α = \frac{ω_{0} - ω}{t} . . . . . . . (i) \\ ω^{2} = ω_{0}^{2} - 2 α θ \\ θ = \frac{(ω_{0} - ω) (ω + ω_{0})}{2 α} = \frac{α t \cdot (ω + \frac{7}{2} ω)}{2 α} = \frac{9}{4} ω t \\ R θ = \frac{9}{4} ω R t . . . . (i i) \\ v = v_{0} + a t (v_{0} = 0) \\ a = ω R / t \\ v^{2} = v_{0}^{2} + 2 a S \\ S = \frac{v \cdot v}{2 a} = \frac{(ω R) (v \cdot t)}{2 \cdot ω R} = \frac{v t}{2} . . . . . . . (i i i) \\ Ratio of distance = 9 / 2 \end{array}$

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