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Paraphin [41]
3 years ago
9

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

he disk. When the disk reaches a speed of 0.8 m/s, the object starts to slide off the disk. What is the coefficient of static friction between the object and the disk?
mass=100 g, r= 0.75 m​
Physics
1 answer:
koban [17]3 years ago
5 0

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (1)

\Sigma F_{y} = N - m\cdot g = 0 (2)

Where:

N - Normal force from the ground on the object, measured in newtons.

m - Mass of the object, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v - Linear speed of rotation of the disk, measured in meters per second.

R - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}

\mu_{s} = \frac{v^{2}}{g\cdot R}

If we know that v = 0.8\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 0.75\,m, then the coefficient of static friction between the object and the disk is:

\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}

\mu_{s} = 0.087

The coefficient of static friction between the object and the disk is 0.087.

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