Problem 6 - College Physics

Question

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the plattorm when it is brought up to a speed of 40.0 \(\mathrm{rev} / \mathrm{min}\) , provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

Step-by-Step Solution

Answer

(a) \(\mu_s = 0.111\); (b) \(r_{max} = 0.098 \) m.

Step 1: Convert Angular Velocity to Rad/Sec

First, convert the angular velocity from revolutions per minute (rev/min) to radians per second (rad/s).- For initial speed: \[\omega_1 = 40.0\, \text{rev/min} \times \left(\frac{2\pi\, \text{rad}}{1\, \text{rev}}\right) \times \left(\frac{1\, \text{min}}{60\, \text{sec}}\right) = \frac{4 \pi}{3} \text{ rad/sec}\]- For final speed:\[\omega_2 = 60.0\, \text{rev/min} \times \left(\frac{2\pi\, \text{rad}}{1\, \text{rev}}\right) \times \left(\frac{1\, \text{min}}{60\, \text{sec}}\right) = 2 \pi \text{ rad/sec}\]

Step 2: Calculate Initial Centripetal Force

The centripetal force keeping the button on the platform is given by: \[F_c = m r \omega_1^2\]Where \(m\) is the mass of the button, \(r\) is the maximum radius (0.150 m), and \(\omega_1\) is the angular velocity in rad/s. Simplifying the equation using the values from Step 1, we get:\(F_c = m \times 0.150 \times \left(\frac{4 \pi}{3}\right)^2\).

Step 3: Static Friction Force Requirement

The force of static friction must be equal to or greater than the centripetal force for the button to stay in place:\[f_s = \mu_s N\]Where \(f_s\) is the force of static friction, \(\mu_s\) is the coefficient of static friction, and \(N = mg\) is the normal force (equal to the button's weight). Since \(f_s = F_c\), we can equate to find:\[\mu_s mg = m \times 0.150 \times \left(\frac{4 \pi}{3}\right)^2\].

Step 4: Solve for Static Friction Coefficient

We eliminate the mass (\(m\)) from both sides since it appears on both sides:\[\mu_s g = 0.150 \times \left(\frac{4 \pi}{3}\right)^2\]Solving for \(\mu_s\):\[\mu_s = \frac{0.150 \times \left(\frac{4 \pi}{3}\right)^2}{9.8}\]\[\mu_s \approx 0.111\].

Step 5: Calculate Maximum Radius for New Speed

To find how far from the axis the button can be placed at 60 rev/min without slipping, we use the friction limit:\[\mu_s mg = m r_{max} \omega_2^2\]Replacing \(\omega_2\) and solving for \(r_{max}\):\[r_{max} = \frac{\mu_s g}{\omega_2^2}\]\[r_{max} = \frac{0.111 \times 9.8}{(2\pi)^2}\]\[r_{max} \approx 0.098\, \text{m}\].

Key Concepts

Angular Velocity

Angular velocity is a critical concept when studying rotational motion. It refers to the rate at which an object rotates or revolves around a center or axis. **In this exercise**, we deal with converting angular velocity from revolutions per minute (rev/min) to radians per second (rad/s). This conversion is essential because radians are the standard unit of angular measure used in physics.

To convert from rev/min to rad/s, you'll follow these steps:
- **Multiply by** the conversion factor from revolutions to radians (\(2\pi \, \text{rad}\)), since one complete revolution is \(2 \pi \) radians.
- **Then, convert** from minutes to seconds by dividing by 60, because there are 60 seconds in a minute.

For example, to convert 40.0 rev/min, the equation becomes:
\[\omega = 40.0 \, \text{rev/min} \times \left(\frac{2\pi \, \text{rad}}{1 \, \text{rev}}\right) \times \left(\frac{1 \, \text{min}}{60 \, \text{sec}}\right)\]
Thus, the angular velocity \(\omega\) is \(\frac{4\pi}{3} \, \text{rad/s}\).

This accurate conversion helps us calculate other forces present on the rotating platform, such as the centripetal force, which keeps an object moving in a circular path.

Static Friction

Static friction is the force that keeps two surfaces from sliding over each other. Instead of moving, it balances the other forces trying to move the object. **In rotational systems**, like the button on a rotating platform, static friction plays a vital role.

**Why is static friction important here?**

• **Centripetal force** is needed to keep the button from slipping off the platform.
• **Static friction** provides that centripetal force. It needs to be at least equal to the force trying to move the object to successfully hold it in place.

For an object to remain at rest relative to the rotating surface, the static friction force \(f_s\) must be greater than or equal to the centripetal force \(F_c\):
\[f_s = \mu_s N\]

Where \(\mu_s\) is the coefficient of static friction, and \(N = mg\) is the normal force (the weight of the object).

**Calculating** the coefficient of static friction in this exercise used:
\[\mu_s g = 0.150 \times \left(\frac{4\pi}{3}\right)^2\]

This tells us how sticky or grippy the surface is. A higher coefficient means more static friction, allowing for more force without slipping.

Rotational Motion

Rotational motion involves objects moving in circular paths around a central point or axis. The principles governing these movements are crucial in understanding many everyday phenomena.

**In this exercise** with the button and rotating platform, several key points of rotational motion come into play:

• **Angular velocity**: This indicates how fast the object is revolving around the axis. A higher angular velocity means faster rotation.
• **Centripetal force**: Required to make an object follow a circular path. Without this force, the object would continue moving in a straight line outward from the center.
• **Static friction**: Provides the necessary centripetal force to keep the button from sliding off, illustrating the critical interaction between linear and rotational forces.

Understanding these concepts helps solve how far from the axis the button can be placed without slipping. We calculate the maximum radius \(r_{max}\) using the formula:
\[r_{max} = \frac{\mu_s g}{\omega_2^2}\]

This allows us to see the intricate balance between the friction, rotational speed, and position that defines rotational motion's stability.