/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 A metal tool is sharpened by bei... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(2.50 \mathrm{rev} / \mathrm{s}\) 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 energy is being transferred at a rate of approximately 565.5 W.

Step by step solution

01

Identify Given Values

We need to define all the given values from the problem statement. These include the force applied on the tool, \(F = 180 \, \text{N}\), the radius of the grinding wheel \(r = 20.0 \, \text{cm} = 0.2 \, \text{m}\), the rotational speed of the wheel \(2.50 \, \text{rev/s}\), and the coefficient of kinetic friction \(\mu_k = 0.320\).
02

Convert Rotational Speed to Radians per Second

The rotational speed needs to be converted to angular velocity in radians per second. The formula for this is \(\omega = 2\pi \times \text{revolutions per second}\). Thus, \(\omega = 2\pi \times 2.50 = 5\pi \, \text{rad/s}\).
03

Calculate Frictional Force

The frictional force \(f\) can be calculated using the formula \(f = \mu_k \times F\). Thus, \(f = 0.320 \times 180 \, \text{N} = 57.6 \, \text{N}\).
04

Calculate Torque

The torque \(\tau\) exerted by the frictional force is \(\tau = r \times f\). Substituting the known values, \(\tau = 0.2 \, \text{m} \times 57.6 \, \text{N} = 11.52 \, \text{Nm}\).
05

Calculate Power Transferred

The power \(P\) is the rate at which work is done and can be calculated using the formula \(P = \tau \times \omega\). Substituting the values, \(P = 11.52 \, \text{Nm} \times 5\pi \, \text{rad/s} = 180\pi \, \text{W}\). This is approximately \(565.5 \, \text{W}\).

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Key Concepts

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

Torque Calculation
Torque is a measure of the rotational force acting on an object. It’s what makes things spin and is calculated using the formula \( \tau = r \times f \). In this case, \( \tau \) represents torque, \( r \) represents the radius at which the force is applied, and \( f \) is the force itself.
In our grinding tool problem, the frictional force \( f \) acting on the tool, calculated as \( f = \mu_k \times F = 0.320 \times 180 \, \text{N} = 57.6 \, \text{N} \), is applied at the rim of the wheel. The radius \( r \) of 0.2 meters means our torque exerted by the frictional force is:
  • \( \tau = 0.2 \, \text{m} \times 57.6 \, \text{N} \)
  • \( \tau = 11.52 \, \text{Nm} \)
Torque helps us understand how efficiently a force can create rotational motion, which in turn affects power and energy transfer from the grinding wheel.
Angular Velocity
Angular velocity measures how fast something spins or rotates, expressed in radians per second. In problems involving rotating objects, such as our grinding wheel, converting linear speed into angular speed helps clarify how quickly the object is turning.
For our wheel, rotating at \(2.50 \, \text{rev/s}\), the angular velocity \( \omega \) is found by:
  • Using the formula \( \omega = 2\pi \times \text{revolutions per second} \)
  • \( \omega = 2\pi \times 2.50 = 5\pi \, \text{rad/s} \)
Understanding angular velocity is crucial for analyzing the power of rotating systems, as it directly influences how much work can be done over time in systems like our grinding machine.
Power Transfer
Power transfer in mechanical systems refers to the rate at which energy is transferred from one part of a system to another. In this context, it involves moving energy from the motor to the grinding wheel, where it becomes thermal energy and kinetic energy.
To find the power being transferred, we use the formula \( P = \tau \times \omega \), where \( P \) is power, \( \tau \) is torque, and \( \omega \) is the angular velocity.
Substituting the values from our exercise:
  • Torque \( \tau = 11.52 \, \text{Nm} \)
  • Angular velocity \( \omega = 5\pi \, \text{rad/s} \)
  • Power \( P = 11.52 \, \text{Nm} \times 5\pi \, \text{rad/s} = 180\pi \, \text{W} \)
  • Approximately \( 565.5 \, \text{W} \)
This power represents how efficiently energy is being converted from electrical to mechanized energy, effectively making the grinding process work on the tool and improve its sharpness.

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Most popular questions from this chapter

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

A machine pulls a \(40 \mathrm{~kg}\) trunk \(2.0 \mathrm{~m}\) up a \(40^{\circ}\) ramp at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is \(0.40 .\) What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

Tarzan, who weighs 688 N, swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long \((\) Fig. \(8-40) .\) From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\). (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. \(8-37\) ) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

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