Question

A coin is placed on a turntable that is rotating at 60. 0 rpm. If the coefficient of static friction between the coin and the turntable is 0. 60, how far from the center of the turntable can the coin be placed without having it slip off?.

Answer 1

The coin can be placed approximately 18.6 cm from the center of the turntable without slipping off.

Step-by-step explanation:

To determine the maximum distance from the center of the turntable that the coin can be placed without slipping off, we need to consider the centripetal force acting on the coin and the maximum static friction force.

Centripetal force = m * r * ω^2, where m is the mass of the coin, r is the distance from the center of the turntable, and ω is the angular velocity in radians per second.

Maximum static friction force = μs * N, where μs is the coefficient of static friction and N is the normal force.

When the coin is at the verge of slipping, the maximum static friction force must equal the centripetal force:

μs * N = m * r * ω^2

Since the normal force N is equal to the weight (mg) of the coin, we can rewrite the equation as:

μs * mg = m * r * ω^2

Now, we can solve for r:

r = (μs * g) / ω^2

Substituting the given values, where the coefficient of static friction (μs) is 0.60 and the angular velocity (ω) is 2π × 60.0 rpm converted to radians per second:

r = (0.60 * 9.8 m/s^2) / (2π × 60.0/60)

r ≈ 0.186 m or 18.6 cm

Answer 2

The coin can be placed at a distance r from the center of the turntable, where: r = (0.60 * g) / (2π)^2

To determine how far from the center of the turntable the coin can be placed without slipping off, 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. The maximum static friction force is given by the equation:

F_static = μ * N

Where F_static is the maximum static friction force, μ is the coefficient of static friction, and N is the normal force.

In this case, the normal force is equal to the weight of the coin, since the coin is not moving vertically. So we have:

N = m * g

Where m is the mass of the coin and g is the acceleration due to gravity.

The centripetal force can be calculated as:

F_c = m * ω^2 * r

Where F_c is the centripetal force, m is the mass of the coin, ω is the angular velocity of the turntable in radians per second, and r is the distance from the center of the turntable to the coin.

Setting the maximum static friction force equal to the centripetal force, we can solve for r:

μ * m * g = m * ω^2 * r

Canceling out the mass m on both sides, we get:

μ * g = ω^2 * r

Solving for r, we find:

r = (μ * g) / ω^2

Given that the turntable is rotating at 60.0 rpm, we need to convert this angular velocity to radians per second:

ω = (60.0 rpm) * (2π rad/1 rev) * (1 min/60 s) = 2π rad/s

Substituting the given values into the equation, we have:

r = (0.60 * g) / (2π)^2

Therefore, the coin can be placed at a distance r from the center of the turntable, where:

r = (0.60 * g) / (2π)^2

It's important to note that the value of g may vary depending on the location on Earth. In this case, g represents the acceleration due to gravity.