Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 9.6 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip

Answer 1

Answer:

(A) Acceleration will be $1.145rad/sec^2$

(B) Coefficient of static friction will be 0.116

Explanation:

We have given angular speed

$\omega =33rpm=\frac{2\pi \times 33}{60}=3.454rad/sec$

Distance from the axis r = 9.6 cm = 0.096 m

(a) Acceleration is equal to

$a_c=\omega ^2r=3.454^2\times 0.096=1.145rad/sec^2$

(b) For seed is not to slip

$ma=\mu mg$

$\mu =\frac{a}{g}$

$\mu =\frac{1.145}{9.8}=0.116$

So coefficient of static friction will be 0.116