91Ó°ÊÓ

A \(50.0\)-kg grindstone is a solid disk 0.520 \(\mathrm{m}\) in diameter. You press an ax down on the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57). The coefficient of kinetion between the blade and the stone is 0.60 , and there is a constant friction torque of 6.50 \(\mathrm{N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle 0.500 \(\mathrm{m}\) long to bring the stone from rest to 120 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (c) How much time does it take the grindstone to come from 120 \(\mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

Step by step solution

01

Convert Angular Speed

First, convert the angular speed from revolutions per minute (rpm) to radians per second (rad/s) for part (a) and (c). Given that 1 revolution = \(2\pi\) radians and 1 minute = 60 seconds.\[\omega = 120 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 4\pi \text{ rad/s}\]

02

Calculate Moment of Inertia

For a solid disk, the moment of inertia \(I\) is given by the formula:\[I = \frac{1}{2} m r^2\]The radius \(r\) is half of the diameter, i.e., \(0.520 \text{ m} / 2 = 0.260 \text{ m}\). Thus,\[I = \frac{1}{2} \times 50.0 \text{ kg} \times (0.260 \text{ m})^2 = 1.69 \text{ kg} \cdot \text{m}^2\]

03

Find Angular Acceleration

Use the formula for angular acceleration \(\alpha\):\[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{4\pi - 0}{9.00 \text{ s}} = \frac{4\pi}{9.00} \text{ rad/s}^2\]

04

Calculate Total Torque Required

The total torque \(\tau_{\text{total}}\) needed is given by:\[\tau_{\text{total}} = I \cdot \alpha + \tau_{\text{friction}} + \tau_{\text{axle}}\]Where \(\tau_{\text{friction}} = f_k \cdot r\) and \(f_k = \mu_k \cdot F_n = 0.60 \times 160 \text{ N}\), so\[f_k = 96 \text{ N}\]Then,\[\tau_{\text{friction}} = 96 \text{ N} \cdot 0.260 \text{ m} = 24.96 \text{ N}\cdot\text{m}\]And the axle friction torque \(\tau_{\text{axle}} = 6.50 \text{ N}\cdot\text{m}\).Thus,\[\tau_{\text{total}} = 1.69 \times \frac{4\pi}{9} + 24.96 + 6.50\]

05

Calculate Force for Part (a)

Rearrange the torque formula \(\tau = r \times F\) to solve for the force \(F\) applied at the crank handle:\[F = \frac{\tau_{\text{total}}}{r} = \frac{1.69 \times \frac{4\pi}{9} + 24.96 + 6.50}{0.500}\]Calculate to find the force.

06

Calculate Force for Part (b)

To maintain constant angular speed, the total torque only needs to counter the friction torques:\[\tau_{\text{steady}} = \tau_{\text{friction}} + \tau_{\text{axle}} = 24.96 + 6.50\]With this torque, use:\[F = \frac{\tau_{\text{steady}}}{r} = \frac{24.96 + 6.50}{0.500}\]

07

Calculate Time for Part (c)

Use the opposite sign for angular acceleration since it is slowing down due to friction:\[\alpha = \frac{\tau_{\text{axle}}}{I} = \frac{6.50}{1.69}\]Using final speed of 0 rad/s:\[t = \frac{\omega_{\text{initial}} - 0}{\alpha} = \frac{4\pi}{\frac{6.50}{1.69}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia

The moment of inertia is essentially the rotational counterpart of mass in linear motion. It acts as a measure of an object's resistance to changes in its rotation. For a solid disk, like the grindstone in our exercise, the moment of inertia (I) is calculated using the equation:\[I = \frac{1}{2} m r^2\]where:

• m is the mass of the object (50.0 kg for the grindstone).
• r is the radius of the disk (0.260 m in this case).

Using these values, the moment of inertia for our grindstone is computed as 1.69 kg·m². This value signifies how the mass is distributed relative to the axis of rotation and directly affects how much effort is needed to spin the stone up to speed or slow it down.

Angular Acceleration

Angular acceleration (\alpha) describes how quickly an object changes its angular velocity. It's similar to the linear acceleration we experience when speeding up or slowing down in a car. In the context of our grindstone,\[\alpha = \frac{\Delta \omega}{\Delta t}\]where:

• \( \Delta \omega \) is the change in angular velocity, computed as the final speed minus the initial speed.
• \( \Delta t \) is the time interval over which this change occurs (9.00 seconds).

For the grindstone problem, moving from 0 to 4\( \pi \) rad/s gives us an angular acceleration of approximately 1.396 rad/s². This quantifies how fast the grindstone is being sped up to reach a particular rotational velocity.

Torque Calculation

Torque is the force that makes an object rotate and is analogous to force in linear dynamics. To calculate the necessary torque to spin up our grindstone, we need the total torque which includes overcoming both friction and providing the necessary acceleration.The total torque (\tau_{ ext{total}}) can be calculated as follows:\[\tau_{ ext{total}} = I \cdot \alpha + \tau_{ ext{friction}} + \tau_{ ext{axle}}\]Here:

• \( I \cdot \alpha \) provides the torque needed for acceleration.
• \( \tau_{\text{friction}} \) is caused by kinetic friction between the axe and stone, found using \( f_k = \mu_k \cdot F_n = 96 \text{ N} \times 0.260 \text{ m} = 24.96 \text{ N}\cdot\text{m} \).
• \( \tau_{\text{axle}} \) is the steady frictional torque of 6.50 \text{ N}\cdot\text{m}.

Once total torque is known, the force applied at the crank can be determined using \( \tau = r \cdot F \).

Kinetic Friction

Kinetic friction is the resistance between two objects sliding past each other. In our situation, this is the friction between the grindstone and the axe, characterized by the coefficient of kinetic friction (\mu_k) of 0.60.The force of kinetic friction (\text{f}_k) is calculated as:\[\text{f}_k = \mu_k \cdot \text{F}_n\]where:

• \( \text{F}_n \) is the normal force, which is 160 N in this exercise.

Multiplying these values gives \( \text{f}_k = 96 \text{ N} \), and when applied over the radius of the grindstone, it contributes significantly to the torque that must be overcome to start or maintain rotation.

Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts on a system, the angular momentum remains constant. For our grindstone, this concept tells us that, without friction or other forces acting, its spinning motion would continue indefinitely once started. In practical terms:

• If only the axle friction acts on the grindstone, the angular momentum will gradually decrease until the stone eventually stops.
• This principle allows us to predict how long it will take for a grindstone to stop if no additional tangential force is applied and only frictional forces act upon it.

Specifically, by calculating how much of the initial angular momentum is lost to frictional forces, we can determine the time it takes to bring the rotating stone to rest by considering only the axle friction's braking effect.