91Ó°ÊÓ

A \(50.0 \mathrm{~kg}\) 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. \(\mathbf{P} 10.58\) ). The coefficient of kinetic friction 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 rev \(/ \min\) in \(9.00 \mathrm{~s} ?\) (b) After the grindstone attains an angular speed of 120 rev \(/ \mathrm{min}\), what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev \(/ \min ?\) (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

Determine the Angular Acceleration

First, we need to convert the angular speed from revolutions per minute to radian per second by multiplying it with \(2\pi/60\) : \(\omega = 120 \ rev/min \times \frac{2\pi}{60} = 4\pi \ rad/s\). Then, using the relation \(\alpha = \omega / t\), the angular acceleration becomes \(\alpha = 4\pi / 9 \ rad/s^2\).

02

Determine the Moment of Inertia

The moment of inertia I of a solid disk is given by the formula \(I = 1/2 \cdot m \cdot r^2\). Substituting the given values, we get \(I = 1/2 \times 50 \times (0.26)^2 = 1.69 \ kg.m^2\).

03

Calculate the total Torque

To find the total torque to accelerate the grindstone, we use the relation \(\sum \tau = I\cdot \alpha + \tau_{friction} + \tau_{blade}\). The torque due to the friction is given as \(6.5 \ N.m\). The torque due to the blade can be calculated using the formula \(\tau_{blade} = F_{blade} \cdot r \cdot \mu_k\) where \(F_{blade} = 160 \ N\), \(r = 0.26 \ m\) and \(\mu_k = 0.6\), which gives \(\tau_{blade} = 160 \times 0.26 \times 0.6 = 24.96 \ N.m\). Now we substitute these calculated and given values to find the total torque.

04

Determine the Force at the Handle to Accelerate Grindstone

The total torque \(\tau_{total}\) needed to achieve the required angular acceleration is the sum of the torques needed to overcome the axle friction, the friction due to axe pressure, and the torque needed to speed up the disk. It can be calculated by \(\sum \tau = I\cdot \alpha + \tau_{friction} + \tau_{blade}\). Substituting the calculated values, we get \(\tau_{total} = 1.69 \times (4\pi/9) + 6.5 + 24.96 = 20.63 \ N.m\). Therefore, the force at the handle needed to spin the grindstone up to speed is \(F = \tau_{total} / r_{crank}\), where \(r_{crank} = 0.5 \ m\). Calculating gives \(F = 20.63 / 0.5 = 41.26 \ N\)

05

Determine the Force at the Handle to Maintain Constant Speed

Once the grindstone is up to speed, no further torque is needed to keep it going, apart from the frictionals torques. Therefore, the total torque needed is equal to the torque of friction of the axle plus the torque due to pressure of the blade, which we have already calculated. The force required at the handle would then be \(F = (\tau_{friction} + \tau_{blade}) / r_{crank}\). Substituting the values gives \(F = (6.5 + 24.96) / 0.5 = 62.92 \ N\)

06

Determine the Time needed for Grindstone to come to Rest

Assuming that the only torque at play when the grindstone comes to rest is due to friction of axle. To calculate the time needed for the grindstone to come to rest, we use the angular speed equation \(\omega = \alpha \ t\). Therefore, \(t = \omega / \alpha\), where \(\alpha\) is the angular deceleration due to the friction torque: \(\alpha = \tau_{friction} / I = 6.5 / 1.69 = 3.85 \ rad/s^2\). Substituting these values gives \(t = 4\pi / 3.85 = 3.26 \ s\)

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Key Concepts

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

Torque Calculation

Torque is essentially the rotational equivalent of linear force. When you apply a force to rotate an object around an axis, you generate torque. It is calculated as the product of the force applied and the distance from the axis:

• Formula: \( \tau = F \times r \)
• \( F \) is the force applied in newtons.
• \( r \) is the distance from the axis of rotation in meters.

In the exercise, we calculated the torque required to overcome various forms of resistance to motion to bring the grindstone up to the desired speed. Each component of the required torque—the frictional torques from the axle and the blade—had to be considered to find the total torque needed.

Moment of Inertia

The Moment of Inertia (MOI) is a measure of an object's resistance to changes in its rotation and is dependent on both its mass and the distribution of that mass relative to the axis of rotation. For a solid disk, the moment of inertia is calculated as:

• Formula: \( I = \frac{1}{2} \cdot m \cdot r^2 \)
• \( m \) is the mass of the object in kilograms.
• \( r \) is the radius of the object in meters.

A high moment of inertia means that it's harder to change the object's rotational state (speed it up, slow it down or change its rotational direction). In the given problem, the moment of inertia helped determine how much torque was needed to accelerate the grindstone from rest.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. In this exercise, it represented how quickly the grindstone could be brought up to its operational speed. It's calculated using the formula:

• Formula: \( \alpha = \frac{\Delta \omega}{\Delta t} \)
• \( \alpha \) is the angular acceleration in radians per second squared.
• \( \omega \) is the angular speed in radians per second.
• \( t \) is the time in seconds during which the change in speed occurs.

We used the angular acceleration to understand the rate at which the grindstone had to be spun up and slowed down based on the forces applied.

Kinetic Friction

Kinetic friction is the resistive force that acts between moving surfaces. When the grindstone is in operation, the ax blade exerts a force on it, and kinetic friction occurs at the point of contact,

• Formula: \( f_k = \mu_k \times F_n \)
• \( \mu_k \) is the coefficient of kinetic friction, which is unitless.
• \( F_n \) is the normal force applied perpendicularly to the contact surface, in newtons.

This frictional force needed to be factored into the total torque to accurately calculate the forces needed to maintain or change the grindstone's speed.

Rotational Dynamics

Rotational dynamics involves the motion of objects rotating about an axis and the factors affecting such motion. It is similar to linear dynamics but in terms of angular quantities, like torque, angular momentum, and angular acceleration. In this particular exercise:

• We analyzed how forces applied tangentially and frictional forces affected the angular acceleration.
• The interplay of forces resulted in changes to the rotational motion of the grindstone.
• Understanding of rotational dynamics was crucial in determining how the grindstone's speed was initialized, maintained, and brought back to rest.

These principles are vital not only for simple mechanical systems like a grindstone but also for understanding complex rotational systems in real-world applications.