Question

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N (Fig. P10.54), and the grindstone comes to rest in 7.50 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Short answer

The coefficient of friction is approximately 0.482.

Step-by-step solution

Convert Angular Velocity to Radians Per Second

First, we need to convert the rotational speed of the grindstone from revolutions per minute to radians per second. 1 revolution is equal to \(2\pi\) radians, and 1 minute is 60 seconds.Given:\[850 \text{ rev/min} = 850 \times \frac{2\pi}{60} \text{ rad/s} \approx 89.012 \text{ rad/s}.\]

Calculate Angular Deceleration

The grindstone comes to a stop, which means its final angular velocity is 0 rad/s. We can use the equation for angular deceleration:\[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 - 89.012}{7.5} = -11.868 \text{ rad/s}^2.\]

Calculate Torque Due to Friction

Torque \(\tau\) caused by friction is given by:\[\tau = I \cdot \alpha,\]where \(I\) is the moment of inertia for a solid disk = \(\frac{1}{2} M R^2\).Given: Diameter = 0.520 m, so \(R = 0.260\,\text{m}\).Mass \(M = 50.0\,\text{kg}\).Calculate \(I\):\[I = \frac{1}{2} \times 50 \times (0.26)^2 = 1.69 \text{ kg m}^2.\]Now calculate \(\tau\):\[\tau = 1.69 \times (-11.868) = -20.03492 \text{ Nm}.\]

Relate Torque to Coefficient of Friction

The torque due to friction is also given by:\[\tau = R \cdot f,\]where \(f\) is the frictional force \(f = \mu N\) (where \(N\) is the normal force).Thus, we have:\[20.03492 = 0.26 \times \mu \times 160.\]Solve for \(\mu\).\[\mu = \frac{20.03492}{0.26 \times 160} \approx 0.482.\]

Conclusion

Based on the calculations, the coefficient of friction between the ax and the grindstone is approximately 0.482.

Key concepts

Angular Deceleration

Angular deceleration is the rate at which an object's rotational speed decreases. Similar to linear deceleration, which involves a decrease in linear speed, angular deceleration is about rotational motion. It's essential in situations where spinning objects come to rest.
To calculate angular deceleration, we use the formula:
\[\alpha = \frac{\Delta \omega}{\Delta t}\]where:

• \(\alpha\) is the angular deceleration.
• \(\Delta \omega\) is the change in angular velocity.
• \(\Delta t\) is the time over which this change occurs.

For example, if a grindstone is initially rotating at 89.012 rad/s and stops in 7.5 seconds, the angular deceleration would be \(-11.868 \text{ rad/s}^2\). The negative sign indicates that it's slowing down.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to angular acceleration, much like mass is a measure of resistance to linear acceleration. It depends not only on the object's mass but also on the distribution of that mass relative to the axis of rotation.
For a solid disk, like the grindstone, the moment of inertia \(I\) is calculated using:
\[I = \frac{1}{2} M R^2\]where:

• \(M\) is the mass of the disk.
• \(R\) is the radius of the disk.

In our case with a grindstone mass of 50 kg and a radius of 0.26 m, the moment of inertia is \(1.69 \text{ kg m}^2\). This value is crucial for calculating torque, as torque is directly related to both the moment of inertia and angular deceleration.

Torque

Torque is the rotational equivalent of force. Just as force causes linear acceleration, torque causes angular acceleration or deceleration. It is calculated as the product of force and the distance from the point of rotation (the lever arm).
However, in rotational dynamics with known moment of inertia and angular deceleration, torque \(\tau\) can be determined using:
\[\tau = I \cdot \alpha\]where \(I\) is the moment of inertia and \(\alpha\) is the angular deceleration.
For the grindstone, where \(I = 1.69 \text{ kg m}^2\) and \(\alpha = -11.868 \text{ rad/s}^2\), the torque exerted by the friction can be calculated as \(-20.03492 \text{ Nm}\). This torque helps us understand the impact friction has in bringing the grindstone to rest.

Coefficient of Friction

The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is represented by \(\mu\) and ranges from 0 (no friction) to 1 or above (very high friction).
It relates to torque in rotational systems, particularly when looking to solve for \(\mu\) between the ax and the grindstone.
Using this relationship:
\[\tau = R \cdot f \quad \text{and} \quad f = \mu N\]where \(N\) is the normal force. We find:
\[\mu = \frac{\tau}{R \cdot N}\]In our scenario, with a torque of \(-20.03492 \text{ Nm}\), a radius \(R = 0.26 \text{ m}\), and a normal force of 160 N, the coefficient of friction is approximately 0.482. This value tells us how effectively the frictional force between the axe and grindstone is applied in slowing down the rotating grindstone.