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

`0.37163\ m/s^2`

0.07576

0.13235

Step-by-step explanation:

`\omega` = Angular speed =
`(100)/(3)* (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* ((100)/(3)* (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=(a_c)/(g)\\\Rightarrow \mu=(0.74326)/(9.81)\\\Rightarrow \mu=0.07576`

The coefficient of friction is 0.07576

Transverse acceleration is given by

`a_t=r(\omega)/(t)`

The resultant acceleration is given by

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

`\mu=(1.29842)/(9.81)\\\Rightarrow \mu=0.13235`

The coefficient of friction is 0.13235