Question

A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is \(0.1,\) how far from the center of the turntable can the coin be placed without having it slip off?

Short answer

Answer: The coin can be placed a maximum distance of approximately 0.0803 m or 8.03 cm from the center of the turntable without slipping off.

Step-by-step solution

Convert rpm to angular velocity

The first thing we need to do is to convert the turntable's rotational speed to angular velocity. This can be done using the following equation: \(\omega = 2\pi*\text{rpm}/\text{60 s}\) Given that the turntable rotates at 33.3 rpm, we can solve for the angular velocity: \(\omega = 2\pi*(33.3\,\text{rpm})/\text{60 s} = 3.49\,\text{rad/s}\)

Set up the centripetal force equation

Next, let's find out the centripetal force act on the coin when it's on the edge of slipping off using the following equation: \(F_c = m(r\omega^2)\) where \(F_c\) is the centripetal force, \(m\) is the mass of the coin, \(r\) denotes the distance from the center of the turntable, and \(\omega\) is the angular velocity.

Set up the static friction force equation

Now, we need to find the maximum static friction force that prevents the coin from slipping off. We can use the static friction coefficient and the normal force to calculate this: \(F_f = \mu F_N\) where \(F_f\) is the friction force, \(\mu\) is the coefficient of static friction, and \(F_N\) is the normal force (equal to the weight of the coin in this case). Since the coin is laying flat on the turntable, we have: \(F_N = mg\) Therefore, the maximum static friction force can be expressed as: \(F_f = \mu mg\)

Equate centripetal force and static friction force

For the coin not to slip off, the centripetal force on the coin should be equal to or less than the maximum static friction force: \(F_c \le F_f\) Using our formulas above, we can write this inequality as: \(m(r\omega^2) \le \mu mg\) The mass of the coin will cancel, leaving: \(r\omega^2 \le \mu g\) Now, we'll solve for the maximum radius (\(r\)) solving for \(r\) in the equation: \(r \le \frac{\mu g}{\omega^2}\)

Calculate the maximum distance (radius)

Let's plug in the numbers to calculate the maximum distance (radius) from the center that coin can be placed without slipping off: \(r \le \frac{0.1\times9.81\,\text{m/s}^2}{(3.49\,\text{rad/s})^2}\) \(r \le 0.0803\,\text{m}\) So, the coin can be placed a maximum distance of approximately \(0.0803\,\text{m}\) or \(8.03\,\text{cm}\) from the center of the turntable without slipping off.