A coin is placed 12.5 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 36 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

a similar word for glare

yes

To find the coefficient of static friction between the coin and the turntable, we can use the concept of centripetal force.

The centripetal force required to keep the coin in circular motion is provided by the static friction between the coin and the turntable. At the point when the coin slides off, the maximum static friction force is reached. Given that the coin slides off when the turntable reaches a rate of 36 rpm, we can convert this to radians per second (ω) by using the formula:

ω = 2π × (rpm) / 60

Substituting the given value, we get:

ω = 2π × 36 / 60 ≈ 3.77 rad/s

The centripetal force required to keep the coin in circular motion is given by the equation:

F = m × ω^2 × r

Where:
F = centripetal force
m = mass of the coin
ω = angular velocity in radians per second
r = distance of the coin from the axis of rotation

Since the coin remains fixed on the turntable until it slides off, the centripetal force is equal to the maximum static friction force (Fs) acting between the coin and the turntable.

So we have:

Fs = m × ω^2 × r

To find the coefficient of static friction (μs), we can rearrange the equation:

Fs = μs × N

Where:
Fs = maximum static friction force
μs = coefficient of static friction
N = normal force

The normal force (N) acting on the coin is equal to the weight of the coin (mg), where g is the acceleration due to gravity.

Combining the equations, we have:

m × ω^2 × r = μs × mg

Simplifying, we get:

μs = (m × ω^2 × r) / (m × g)

The mass of the coin cancels out, giving us:

μs = ω^2 × r / g

Now we can plug in the given values:
ω = 3.77 rad/s
r = 12.5 cm = 0.125 m
g ≈ 9.8 m/s^2 (approximate acceleration due to gravity)

μs = (3.77^2 × 0.125) / 9.8 ≈ 0.058

Therefore, the coefficient of static friction between the coin and the turntable is approximately 0.058.