First slide

Rolling motion

Question

A plank P is moving with constant velocity V on a horizontal surface. A disc of radius R is rolling without slipping on the plank with a velocity V relative to the plank. Then radius of curvature of the trajectory of a point on the circumference of the ring when it is at position A, as viewed by an observer standing on the ground is

Difficult

A

8R
B

5R
C

9R
D

4R

Solution

As viewed by an observer on the ground, velocity of A is given by
${\vec{V}}_{A} = {\vec{V}}_{C} + {\vec{V}}_{A / C} = (V + V) \hat{i} + ωR \hat{j} = (2 v \hat{i} + v \hat{j})$
$∴ V_{A} = \sqrt{{(2 v)}^{2} + v^{2}} = \sqrt{5} V$
Now acceleration of A is given by
${\vec{a}}_{A} = {\vec{a}}_{C} + {\vec{a}}_{A / C} = 0 + (ω^{2} R) i = \frac{v^{2}}{R} \hat{i}$
$∴ \frac{V_{A}^{2}}{ρ} = a_{A} \Rightarrow \frac{{(\sqrt{5} v)}^{2}}{ρ} = \frac{v^{2}}{R} \Rightarrow ρ = 5 R$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App