Problem 59 - University Physics with Modern Physics

Question

A grindstone in the shape of a solid disk with diameter 0.520 \(\mathrm{m}\) and a mass of 50.0 \(\mathrm{kg}\) is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57), 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.

Step-by-Step Solution

Answer

The coefficient of friction is approximately 0.482.

Step 1: Convert Rotational Speed to Radians per Second

Since the grindstone's speed is given in revolutions per minute (rev/min), we first convert it to radians per second. One revolution is equal to \(2\pi\) radians. Therefore, \[ \text{Initial angular speed} \ \omega_i = 850 \times \frac{2\pi \text{ radians}}{60 \text{ seconds}}. \]Calculating this gives \[ \omega_i \approx 88.96 \text{ rad/s}. \]

Step 2: Calculate Deceleration Using Angular Kinematics

The final angular speed \(\omega_f\) is 0 rad/s because the grindstone comes to rest. Using the formula \[ \omega_f = \omega_i + \alpha t, \]we solve for angular deceleration \(\alpha\):\[ 0 = 88.96 + \alpha \times 7.50, \]\[ \alpha \approx -11.86 \text{ rad/s}^2. \]

Step 3: Determine Torque Required to Stop the Grindstone

The torque \(\tau\) is related to angular deceleration by the equation \[ \tau = I \alpha, \]where \(I\) is the moment of inertia of the grindstone. For a solid disk, \[ I = \frac{1}{2}mr^2, \]where \(m\) is mass and \(r\) is radius. Given \(m = 50.0 \text{ kg}\) and \(r = 0.260 \text{ m}\),\[ I = \frac{1}{2} \times 50.0 \times (0.260)^2 = 1.69 \text{ kg} \cdot \text{m}^2. \]Thus, \[ \tau = 1.69 \times (-11.86) = -20.05 \text{ Nm}. \]

Step 4: Relate Friction Force and Torque

The frictional force \(f\) is related to the torque by \[ \tau = f \times r. \]So, \[ f = \frac{-20.05}{0.260} \approx 77.12 \text{ N}. \]

Step 5: Calculate the Coefficient of Friction

The coefficient of friction \(\mu\) is calculated using \[ f = \mu \times N, \]where \(N\) is the normal force. Therefore, \[ 77.12 = \mu \times 160, \]\[ \mu = \frac{77.12}{160} \approx 0.482. \]

Key Concepts

Grindstone

A grindstone, typically shaped as a solid disk, is a vital tool for sharpening and smoothing edges. It spins rapidly, creating friction when an object is pressed against its surface. This friction is utilized to grind materials, such as wood or metal, effectively.

In our exercise, the grindstone has a diameter of 0.520 meters and a mass of 50.0 kilograms. Its shape and mass distribution are crucial in calculating its moment of inertia, which is a measure of how much torque is needed for it to rotate around its axis.

• Diameter impacts the radius, determining the rotational distance.
• Mass affects the moment of inertia, which is essential in understanding the rotational dynamics.

Understanding these properties helps in analyzing the rotational motion and forces applied to the grindstone.

Torque

Torque is a measure of the force that causes an object to rotate around an axis. It is the rotational equivalent of linear force. Mathematically, torque (\( \tau \)) is defined as the product of the force applied and the radius at which the force is applied (\( r \)).

In the exercise, torque is a key component as it relates to the angular deceleration of the grindstone. We calculate the torque required to stop the grindstone by using the grindstone's moment of inertia (\( I \)) and its angular deceleration (\( \alpha \)). The formula used is:

• \( \tau = I \alpha \)

This equation shows that torque is directly proportional to angular deceleration, highlighting the resistance to rotational change. The negative value indicates that the torque acts in the opposite direction to the initial motion, working to slow it down.

Angular Deceleration

Angular deceleration refers to the rate at which an object's angular velocity decreases over time. It is an essential aspect of rotational motion, comparable to linear deceleration in straight-line motion.

To find angular deceleration in our exercise, we use the equation that links initial and final angular speeds to deceleration over time:

• \( \omega_f = \omega_i + \alpha t \)

In this equation, \( \omega_i \) represents the initial angular speed, \( \alpha \) is the angular deceleration, and \( t \) is time. With an initial speed of 88.96 rad/s and coming to a rest in 7.50 seconds, we calculate an angular deceleration of -11.86 rad/s².

This negative value indicates that the grindstone's rotation is slowing down, a process directly influenced by the applied torque and friction.

Coefficient of Friction

The coefficient of friction (\( \mu \)) quantifies the amount of frictional resistance between two surfaces. It varies based on material properties and surface roughness. The coefficient has no units because it represents a ratio of the frictional force (\( f \)) to the normal force (\( N \)).

In this exercise, we needed to find the coefficient of friction between the ax and the grindstone as they interact.

• Frictional Force: Calculated using torque and radius, \( f = \frac{\text{Torque}}{r} \).
• Normal Force: The force exerted perpendicular to the surfaces in contact, \( N = 160 \text{ N} \).

By rearranging the equation \( f = \mu N \), we solve for \( \mu \) and find it to be approximately 0.482.

Understanding this concept is crucial when analyzing the effects of frictional forces in rotational and linear dynamics.