Question

A ring of radius R is rolling purely on the outer surface of a fixed pipe of radius 4R. At some instant, the centre of the ring has a constant speed = v as shown in fig.

• A. The acceleration of the point on the ring which is in contact with the surface of the pipe w.r.t ground is $\frac{4v^2}{5R}$
• B. The acceleration of the point on the ring which is in contact with the surface of the pipe w.r.t ground is $\frac{v^2}{R}$
• C. The acceleration of the point on the ring which is farthest from the centre of the pipe at the given moment is $\frac{6v^2}{5R}$
• D. Acceleration of B w.r.t. A is $\frac{v^2}{R}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

The acceleration of the point on the ring which is in contact with the surface of the pipe w.r.t ground is $\frac{4v^2}{5R}$

C

The acceleration of the point on the ring which is farthest from the centre of the pipe at the given moment is $\frac{6v^2}{5R}$

D

Acceleration of B w.r.t. A is $\frac{v^2}{R}$

$|{\overrightarrow{a}}_{AC}|=\frac{v^2}{5R}or{\overrightarrow{a}}_{BA}={\overrightarrow{a}}_{BC}-{\overrightarrow{a}}_{AC}or|{\overrightarrow{a}}_{BA}|=\frac{v^2}{R}and{\overrightarrow{a}}_{BC}={\overrightarrow{a}}_{BA}+{\overrightarrow{a}}_{AC}or{\overrightarrow{a}}_{BC}=\frac{v^2}{R}(1-\frac{1}{5})=\frac{4v^2}{5R};|{\overrightarrow{a}}_{DC}|=\frac{v^2}{R}+\frac{v^2}{5R}=\frac{6v^2}{5R}$