Question

A coin kept at a distance of 5 cm from the centre of a turntable of radius 1.5 m

Answer 1

Answer:

Diameter is 3cm and area is

Answer 2

Complete Question:

A Coin kept at a distance of 5 cm from the centre of a turntable of radius 1.5 m just begins to slip when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable. [g= 9.8 m/s^2]

Answer:

The coefficient of static friction between the coin and the turntable is 0.4527

Given:

Distance of coin from centre = r = 5 cm = 0.05 m

Speed of rotation = n = 90 r.p.m = 1.5 Hz = Linear Frequency

To find:

Coefficient of static friction = ?

Formula used:

Limiting force of static friction = Centrifugal force

$\bold{\mu_s mg=mr\omega^{2}}$

Where,

$\bold{\mu_s}$ = Coefficient of static friction

m = Mass

g = Acceleration due to gravity

$\bold{\omega}$ = Angular frequency (Not given)

$\bold{\omega=2\pi n}$

n = Linear Frequency

Solution:

Coefficient of static friction:

$\bold{\mu_s =\frac{mr\omega^{2}}{mg}}$

$\bold{\mu_s=\frac{r\omega^2}{g}}$

$\bold{\mu_s=\frac{r(2\pi n)^2}{g}}$

$\bold{\mu_s=\frac{0.05\times4\times\pi^2\times(1.5)^2}{9.8}}$

$\bold{\therefore\mu_s=0.4527}$