Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 5.1 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) The acceleration is 0.608 $\frac{m}{s^{2} }$

(b) The minimum value of coefficient of static friction is 0.062

Explanation:

Given:

Angular frequency $\omega = 33 \times \frac{2\pi }{60} = 3.454$ $\frac{rad}{s}$

Distance between axis of rotation and watermelon $r = 5.1 \times 10^{-2}$ m

(a)

For calculating acceleration,

From the formula of centripetal acceleration,

  $a_{c} = r \omega ^{2}$

  $a_{c} = 5.1 \times 10^{-2} \times (3.454) ^{2}$

  $a_{c} = 0.608$ $\frac{m}{s^{2} }$

(b)

For finding the minimum value of the coefficient of static friction,

  $F = ma_{c}$

But here only friction force act

  $\mu_{s} mg = ma_{c}$

Where $g = 9.8 \frac{m}{s^{2} }$

  $\mu _{s }= \frac{a_{c} }{g}$

  $\mu _{s} = \frac{0.608}{9.8}$

  $\mu _{s} = 0.062$

Therefore, the acceleration is 0.608 $\frac{m}{s^{2} }$ and minimum value of coefficient of static friction is 0.062