Question

A disc revolves with a speed $33 \frac { 1 } { 3 }$ rev/ min, and has a radius of 15 cm. Two coins A and B are placed at 4 cm and 14 cm away from the centre of the disc. If the coefficient of friction between the coins and the disc is 0.15, which of the coins will revolve with the record?

• A. A
• B. B
• C. Both A and B
• D. Neither A nor B

Answer 1

Answer: (A)

A

Answer 2

Here, $\mu = 0.15$

$\mathrm { v } = 33 \frac { 1 } { 3 } r p m = \frac { 100 } { 3 } r p m = \frac { 100 } { 3 \times 60 } r p s = \frac { 5 } { 9 } r p s$

$\therefore \omega = 2 \pi \mathrm { v } = 2 \times \frac { 22 } { 7 } \times \frac { 5 } { 9 } = \frac { 220 } { 63 } \mathrm { rads } ^ { - 1 }$

The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force

i.e. $\frac { m v ^ { 2 } } { r } \leq \mu m g$

As $v = r \omega \therefore m \omega ^ { 2 } r \leq \mu m g$

For the given $\mu$ and $\omega$ the condition is

$r \leq \frac { \mu g } { \omega ^ { 2 } } \ldots \ldots ( i )$ $A s \frac { \mu g } { \omega ^ { 2 } } = \frac { 0.15 \times 10 } { \left( \frac { 220 } { 63 } \right) ^ { 2 } } \approx 12 \mathrm {~cm}$

For coin A , r=4 cm

The above condition is satisfied, therefore coin A will revolve with the disc.

For coin B, r = 14cm

The above condition Is not satisfied, therefore coin B will not revolve with the disc

Note: We have noting to do with the radius of the disc.