Physics, 05.05.2020 23:35 junkmailemail42

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.

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Answer from: Jasten

a) $a_{r} = 0.275\,\frac{m}{s^{2}}$ , b) $\mu_{s} = 0.028$ , c) $\mu_{s} = 0.036$

Explanation:

a) The linear acceleration of the watermelon seed is:

$a_{r} = \omega^{2}\cdot r$

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

$a_{r} = 0.275\,\frac{m}{s^{2}}$

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:

$\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a$

$a = \mu_{s}\cdot g$

$\mu_{s} = \frac{a}{g}$

$\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$\mu_{s} = 0.028$

c) Angular acceleration experimented by the turntable is:

$\alpha = \frac{\omega-\omega_{o}}{\Delta t}$

$\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}$

$\alpha = 9.6\,\frac{rad}{s^{2}}$

The tangential acceleration experimented by the watermelon seed is:

$a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)$

$a_{t} = 0.221\,\frac{m}{s^{2}}$

The linear acceleration experimented by the watermelon seed is:

$a = \sqrt{a_{t}^{2}+a_{r}^{2}}$

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

$a = 0.353\,\frac{m}{s^{2}}$

The minimum coefficient of static friction is:

$\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$\mu_{s} = 0.036$

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

None. nothing. zero. zilch.

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the...