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Chapter 9: Problem 38

A grinding wheel in the shape of a solid disk is 0.200 m in diameter and has a mass of 3.00 kg. The wheel is rotating at 2200 rpm about an axis through its center. (a) What is its kinetic energy? (b) How far would it have to drop in free fall to acquire the same amount of kinetic energy?

Short Answer

Expert verified
(a) 398.43 J; (b) 13.55 m.

Step by step solution

01

Calculate the Moment of Inertia

The moment of inertia \(I\) for a solid disk rotating about its center is given by the formula \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius. First, convert the diameter to radius: \( r = \frac{0.200\, \text{m}}{2} = 0.100\, \text{m} \). Now, substitute in the values: \[ I = \frac{1}{2} \times 3.00\, \text{kg} \times (0.100\, \text{m})^2 = 0.015\, \text{kg}\,\text{m}^2. \]
02

Convert Rotational Speed to Radians per Second

The rotational speed \( \omega \) must be in radians per second. Convert \( \text{rpm} \) to \( \text{radians per second} \) using the conversion factor \( \frac{2\pi}{60} \). So, \( \omega = 2200\, \text{rpm} \times \frac{2\pi}{60} = \frac{2200 \times 2\pi}{60}\, \text{rad/s}. \) Simplifying gives \( \omega \approx 230.4\, \text{rad/s}. \)
03

Calculate Rotational Kinetic Energy

The rotational kinetic energy \( K \) is given by \( K = \frac{1}{2} I \omega^2 \). Substitute the previously calculated values into this equation: \[ K = \frac{1}{2} \times 0.015\, \text{kg}\,\text{m}^2 \times (230.4\, \text{rad/s})^2. \] This gives \( K \approx 398.43\, \text{J} \).
04

Calculate Equivalent Free Fall Height

The potential energy \( U \) in free fall is given by \( U = mgh \), where \( g = 9.81\, \text{m/s}^2 \) is the gravitational acceleration and \( h \) is the height. Set this equal to the kinetic energy: \( 398.43\, \text{J} = 3.00\, \text{kg} \times 9.81\, \text{m/s}^2 \times h \). Solve for \( h \): \[ h = \frac{398.43\, \text{J}}{3.00\, \text{kg} \times 9.81\, \text{m/s}^2} \approx 13.55\, \text{m}. \]

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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 an essential concept in rotational dynamics. Think of it as the rotational analog of mass in linear motion. It helps determine how difficult it is to change the rotational motion of an object. For a solid disk, like our grinding wheel, the moment of inertia is calculated using the formula: \[ I = \frac{1}{2} m r^2 \] where
  • \( m \) represents the mass of the object,
  • \( r \) is the radius of the disk.
It's crucial to convert the diameter of the disk into the radius by dividing by two. In this example: \[ r = \frac{0.200\, \text{m}}{2} = 0.100\, \text{m} \]The moment of inertia gives insight into the object's resistance to change in its angular speed, and it's necessary for calculating rotational kinetic energy.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins. It indicates the angle an object sweeps out per unit of time, much like speed in linear motion. However, in rotational dynamics, we express this in radians per second.To convert from revolutions per minute (rpm) to radians per second, use the formula:\[ \omega = \text{rpm} \times \frac{2\pi}{60} \]For the grinding wheel, with a speed of 2200 rpm, the conversion looks like this:\[ \omega = 2200 \times \frac{2\pi}{60} = 230.4\, \text{rad/s} \]Understanding angular velocity helps in determining the rotational kinetic energy since it reflects the speed at which the object is spinning.
Potential Energy Gravitational
Gravitational potential energy is the energy stored due to an object's position relative to Earth. It's the energy an object possesses because of its height above the ground. The formula used is:\[ U = mgh \]where
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (9.81 m/s虏),
  • \( h \) is the height above the ground.
In the problem, you equate this potential energy to the rotation's kinetic energy to find how far the grinding wheel would need to drop to gain the same energy. Knowing gravitational potential energy helps connect concepts of mechanics, showing how energy transforms from one type to another while remaining conserved.

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Most popular questions from this chapter

A string is wrapped several times around the rim of a small hoop with a radius of 0.0800 \(\mathrm{m}\) and a mass of 0.180 \(\mathrm{kg}\) . If the free end of the string is held in place and the hoop is released from rest (see Figure 9.30), calculate the angular speed of the rotating hoop after it has descended 0.750 \(\mathrm{m} .\)

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm v (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

Storing energy in flywheels. It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large fly wheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius 0.500 \(\mathrm{m}\) and outer radius \(1.50 \mathrm{m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) (a) If, for stability, such a heavy flywheel is limited to 1.75 second for each revolution and has negligible friction at its axle, what must be its length to store 2.5 \(\mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

\(\bullet\) Ultracentrifuge. Find the required angular speed (in rpm) of an ultracentrifuge for the radial acceleration of a point 2.50 \(\mathrm{cm}\) from the axis to equal \(400,000 g .\)

A \(\mathrm{A} 392\) -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 J. Calculate \(h .\)
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