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Chapter 8: Problem 98

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of \(180 \mathrm{~N}\). The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of \(20.0 \mathrm{~cm}\) and rotates at \(2.50\) rev/s. The coefficient of kinetic friction between the wheel and the tool is \(0.320 .\) At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

Short Answer

Expert verified
The rate of energy transfer is approximately 181 W.

Step by step solution

01

Calculate the angular velocity

First, convert the rotational speed from revolutions per second to radians per second. Since there are \(2\pi\) radians in a revolution, the angular velocity \(\omega\) is:\[\omega = 2.50 \text{ rev/s} \times 2\pi \text{ rad/rev} = 5\pi \text{ rad/s}\]
02

Determine the frictional force

The frictional force can be found using the formula:\[F_{\text{friction}} = \mu_k \times F_N\]where \(\mu_k = 0.320\) is the coefficient of kinetic friction and \(F_N = 180 \text{ N}\) is the normal force. So,\[F_{\text{friction}} = 0.320 \times 180 \, \text{N} = 57.6 \, \text{N}\]
03

Compute the power transferred

The rate of energy transfer (Power) due to friction is given by the formula:\[P = F_{\text{friction}} \times \text{v}\]where \(\text{v}\) is the linear velocity of the rim of the wheel, calculated as:\[\text{v} = \omega \times r\]where \(\omega = 5\pi \text{ rad/s}\) and \(r = 0.20 \text{ m}\). So:\[\text{v} = 5\pi \text{ m/s} \times 0.20 \, \text{m} = \pi \, \text{m/s}\]Then the power is:\[P = 57.6 \, \text{N} \times \pi \, \text{m/s} \approx 180.85 \, \text{W}\]
04

Result

The rate at which energy is being transferred from the motor to the thermal energy and kinetic energy of the material is approximately 180.85 W.

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Key Concepts

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

Kinetic Friction
Kinetic friction plays a crucial role when a metal tool is being sharpened on a grinding machine. This resistance force occurs when two surfaces are in motion relative to each other. In our exercise, the grinding wheel's surface and the metal tool's edge generate kinetic friction as they rub against one another.
To understand kinetic friction, we need to consider the normal force and the coefficient of kinetic friction.
  • The normal force is the force acting perpendicular to the surfaces in contact. Here, the normal force exerted on the tool is 180 N.
  • The coefficient of kinetic friction ( \(\mu_k = 0.320\)) is a dimensionless value representing how much two surfaces resist motion. Every set of materials has its unique value.
With these components, the frictional force can be calculated using the equation:
\[ F_{\text{friction}} = \mu_k \times F_N \]
Thus, the frictional force in this scenario is 57.6 N. This value represents the resistance the tool faces as it moves over the wheel, directly impacting how efficiently the tool gets sharpened.
Power Transfer
Power transfer refers to the rate at which energy is moved from one system to another. In this exercise, energy supplied by the motor of the grinding machine is transferred into both thermal energy and kinetic energy during sharpening. As the grinding wheel rotates, it transfers power to the tool via frictional forces.
Power can be calculated by multiplying the friction force by the linear velocity of the grinding wheel:
  • The linear velocity of the wheel describes how quickly a specific point on its surface moves. It can be found by multiplying the angular speed (in rad/s) by the radius of the wheel (in meters).
  • The formula: \[ P = F_{\text{friction}} \times \text{v} \], where \( \text{v} \) is linear velocity.
In our case, the linear velocity is derived from the wheel's angular speed of \(5\pi\) rad/s and a radius of 0.20 m, resulting in a \(\pi\) m/s speed. The power transferred, or the rate of energy conversion, is calculated to be approximately 180.85 watts. This energy transfer rate affects how quickly the metal tool is worn down and sharpened.
Energy Conversion
Energy conversion in our grinding machine involves changing electrical energy into mechanical energy, which subsequently transforms into thermal and kinetic energy. The motor initially draws electrical power, which it converts into mechanical power to rotate the wheel at a constant speed. This mechanical energy conversion is crucial to the grinding process.
During the grinding process:
  • Thermal energy is generated due to the friction between the wheel and the tool. This friction heats up both objects, which is one of the reasons the tool material gets removed.
  • Kinetic energy is present as the material gets dislodged from the tool. The high-speed rotation of the wheel throws these particles out, using the finely honed kinetic energy.
This simultaneous conversion process ensures that the tool material is efficiently ground down. The grinding wheel must maintain optimal angular velocity to sustain these energy conversions effectively. Understanding energy conversion helps in evaluating the efficiency and effectiveness of the grinding machine in sharpening tools.

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Most popular questions from this chapter

The pulley has negligible mass, and both it and the inclined plane are frictionless. Block \(A\) has a mass of \(1.0\) \(\mathrm{kg}\), block \(B\) has a mass of \(2.0 \mathrm{~kg}\), and angle \(\theta\) is \(30^{\circ} .\) If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block \(B\) has fallen \(25 \mathrm{~cm} ?\)

A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction between the trunk and the incline is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

A volcanic ash flow is moving across horizontal ground when it encounters a \(10^{\circ}\) upslope. The front of the flow then travels 920 \(\mathrm{m}\) up the slope before stopping. Assume that the gases entrapped in the flow lift the flow and thus make the frictional force from the ground negligible; assume also that the mechanical energy of the front of the flow is conserved. What was the initial speed of the front of the flow?

We move a particle along an \(x\) axis, first outward from \(x=1.0\) \(\mathrm{m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m}\), while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for situations, where \(x\) is in meters: $$\begin{array}{ll} \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array}$$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?
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