Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 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? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.36 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Answer 1

a)
`a_(r) = 0.275\,(m)/(s^(2))`, b)
`\mu_(s) = 0.028`, c)
`\mu_(s) = 0.036`

Step-by-step explanation:

a) The linear acceleration of the watermelon seed is:

`a_(r) = \omega^(2)\cdot r`

`a_(r) = \left[\left(33\,(rev)/(min) \right)\cdot \left(2\pi\,(rad)/(rev) \right)\cdot \left((1)/(60)\,(min)/(s) \right)\right]^(2)\cdot (0.023\,m)`

`a_(r) = 0.275\,(m)/(s^(2))`

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

`\Sigma F = \mu_(s)\cdot m\cdot g = m\cdot a`

`a = \mu_(s)\cdot g`

`\mu_(s) = (a)/(g)`

`\mu_(s) = (0.275\,(m)/(s^(2)) )/(9.807\,(m)/(s^(2)) )`

`\mu_(s) = 0.028`

c) Angular acceleration experimented by the turntable is:

`\alpha = (\omega-\omega_(o))/(\Delta t)`

`\alpha = (3.456\,(rad)/(s)-0\,(rad)/(s) )/(0.36\,s)`

`\alpha = 9.6\,(rad)/(s^(2))`

The tangential acceleration experimented by the watermelon seed is:

`a_(t) = \left(9.6\,(rad)/(s^(2)) \right)\cdot (0.023\,m)`

`a_(t) = 0.221\,(m)/(s^(2))`

The linear acceleration experimented by the watermelon seed is:

`a = \sqrt{a_(t)^(2)+a_(r)^(2)}`

`a = \sqrt{\left(0.221\,(m)/(s^(2)) \right)^(2)+\left(0.275\,(m)/(s^(2)) \right)^(2)}`

`a = 0.353\,(m)/(s^(2))`

The minimum coefficient of static friction is:

`\mu_(s) = (0.353\,(m)/(s^(2)) )/(9.807\,(m)/(s^(2)) )`

`\mu_(s) = 0.036`