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Chapter 10: Problem 64

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Short Answer

Expert verified
(a) 0.15 kg路m虏; (b) 3.615 rad/s.

Step by step solution

01

Calculate Rotational Inertia using the Parallel Axis Theorem

The rotational inertia of a solid cylinder about an axis parallel to its central axis at a distance \( d \) is given by the parallel axis theorem:\[I = I_{cm} + md^2\]where \( I_{cm} = \frac{1}{2}mR^2 \) is the moment of inertia about the center mass axis, \(m\) is the mass, \(d\) is the distance from the central axis to the given axis, and \(R\) is the radius of the cylinder.Substituting the values:\( m = 20 \, \text{kg}, R = 0.1 \, \text{m}, d = 0.05 \, \text{m} \):\[I_{cm} = \frac{1}{2} \times 20 \times (0.1)^2 = 0.1 \, \text{kg} \, \text{m}^2\]Now, calculate \( I \):\[I = 0.1 + 20 \times (0.05)^2 = 0.15 \, \text{kg} \, \text{m}^2\]
02

Use Energy Conservation to Find Angular Speed

When the cylinder is released from its starting position, its gravitational potential energy is converted entirely into rotational kinetic energy at the lowest position.The potential energy at the initial position:\[PE = mgd\]where \( g = 9.8 \, \text{m/s}^2 \).The rotational kinetic energy at the lowest point:\[KE = \frac{1}{2}I\omega^2\]At the lowest point, potential energy equals the rotational kinetic energy:\[mgd = \frac{1}{2}I\omega^2\]Solve for \( \omega \):\[\omega = \sqrt{\frac{2mgd}{I}}\]Substitute the known values:\[\omega = \sqrt{\frac{2 \times 20 \times 9.8 \times 0.05}{0.15}} = \sqrt{13.0667} \approx 3.615 \, \text{rad/s}\]

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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, often symbolized as \(I\), is a measure of an object's resistance to changes in its rotational motion, similar to mass in linear motion. For a solid cylinder, the moment of inertia about its central axis can be calculated using the formula:
  • \[ I_{cm} = \frac{1}{2} m R^2 \]
where:
  • \(m\) is the mass of the cylinder.
  • \(R\) is the radius of the cylinder.
In the exercise, the cylinder's mass is \(20 \, \text{kg}\) and the radius is \(0.1 \, \text{m}\). Substituting these values gives \(I_{cm} = 0.1 \, \text{kg} \, \text{m}^2\). Understanding the concept of moment of inertia is crucial for solving problems involving rotational dynamics, as it influences how easily an object can be set into rotational motion.
Parallel Axis Theorem
The parallel axis theorem allows us to calculate the moment of inertia of an object about any axis that is parallel to an axis through the center of mass. It is expressed by the formula:
  • \[ I = I_{cm} + m d^2 \]
where:
  • \( I_{cm} \) is the moment of inertia about the center of mass axis.
  • \( m \) is the mass of the object.
  • \( d \) is the perpendicular distance between the two axes.
In the problem, since the rotation axis is \(5.0 \text{ cm} or 0.05 \, \text{m}\) from the center axis, we calculate:
  • \( I = 0.1 + 20 \times (0.05)^2 = 0.15 \text{ kg} \, \text{m}^2 \)
This theorem is very useful when dealing with complex systems where the axis of rotation does not coincide with the center of mass, simplifying many practical problems in rotational motion.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotational motion. It is defined by the equation:
  • \[ KE = \frac{1}{2} I \omega^2 \]
where:
  • \( I \) is the moment of inertia.
  • \( \omega \) is the angular speed of the object.
In our scenario, when the cylinder is released from rest, it initially has gravitational potential energy, which gets converted into rotational kinetic energy as it reaches its lowest point. This conversion makes it possible to find the angular speed, \( \omega\), essential for determining how fast the cylinder spins at the lowest point of its motion.Understanding rotational kinetic energy is key to solving problems where objects rotate or spin and helps explain how energy transformation occurs in rotational systems.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed; it can only be changed from one form to another. In the context of rotational mechanics, this means that in an isolated system, the total energy remains constant. In this exercise:
  • The gravitational potential energy at the cylinder's initial height is transformed into rotational kinetic energy as it falls.
The formula representing this conservation is:
  • \[ mgd = \frac{1}{2} I \omega^2 \]
where:
  • \( m \) is the mass of the cylinder.
  • \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
  • \( d \) is the height difference, equivalent to the distance from the central axis.
  • \( \omega \) is the angular velocity.
By using the conservation of energy in this rotational context, you can deduce the velocity and energy transformations involved, ensuring a thorough understanding of the mechanics in play.

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Most popular questions from this chapter

A disk, with a radius of \(0.25 \mathrm{~m},\) is to be rotated like a merry-go- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first 400 rad and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s} .\) Calculate \((\mathrm{a})\) the angular acceleration, \((\mathrm{b})\) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev \(/\) s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

A bicyclist of mass \(70 \mathrm{~kg}\) puts all his mass on each downwardmoving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be \(0.40 \mathrm{~m},\) and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length \(55.0 \mathrm{~m}\) At the instant it makes an angle of \(35.0^{\circ}\) with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle \(\theta\) is the tangential acceleration equal to \(g ?\)

A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev \(/\) min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 1000 rev/min in 60.0 s? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s} ?\)
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