Question

A record turntable is rotating at 33 1/3 rev/min. A watermelon seed is on the turntable 6.1 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. .7432 Correct: Your answer is correct. m/s2 (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? .0758 Correct: Your answer is correct. (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.20 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Answer 1

Answer:

$0.37163\ m/s^2$

0.07576

0.13235

Explanation:

$\omega$ = Angular speed = $\dfrac{100}{3}\times \dfrac{2\pi}{60}$

r = Distance from the center = 6.1 cm

g = Acceleration due to gravity = 9.81 m/s²

Acceleration of the seed would be

$a_c=r\omega^2\\\Rightarrow a_c=0.061\times (\dfrac{100}{3}\times \dfrac{2\pi}{60})^2\\\Rightarrow a_c=0.74326\ m/s^2$

The acceleration of the seed is $0.74326\ m/s^2$

Frictional force is given by

$f=\mu mg\\\Rightarrow ma_c=\mu mg\\\Rightarrow \mu=\dfrac{a_c}{g}\\\Rightarrow \mu=\dfrac{0.74326}{9.81}\\\Rightarrow \mu=0.07576$

The coefficient of friction is 0.07576

Transverse acceleration is given by

$a_t=r\dfrac{\omega}{t}$

The resultant acceleration is given by

$a=\sqrt{a_c^2+a_t^2}\\\Rightarrow a=\sqrt{0.74326^2+(0.061\times \dfrac{100}{3}\times \dfrac{2\pi}{60}\times \dfrac{1}{0.2})^2}\\\Rightarrow a=1.29842\ m/s^2$

$\mu=\dfrac{1.29842}{9.81}\\\Rightarrow \mu=0.13235$

The coefficient of friction is 0.13235