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Chapter 10: Problem 35

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at \(235 \mathrm{rad} / \mathrm{s}\), have the same mass of \(1.25 \mathrm{~kg}\) but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius \(0.25 \mathrm{~m}\), and \((\mathrm{b})\) the larger cylinder, of radius \(0.75 \mathrm{~m} ?\)

Short Answer

Expert verified
The rotational kinetic energies are 1080.5 J for the smaller and 9724.5 J for the larger cylinder.

Step by step solution

01

Understanding Rotational Kinetic Energy

Rotational kinetic energy for a rotating object can be calculated using the formula: \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
02

Calculate Moment of Inertia for a Cylinder

The moment of inertia \( I \) for a solid cylinder rotating about its central axis is given by \( I = \frac{1}{2} m r^2 \). We will use this formula to find the inertia for both cylinders.
03

Determine Moment of Inertia for the Smaller Cylinder

Substitute the given values for the smaller cylinder into the inertia formula: \( m = 1.25 \mathrm{~kg} \) and \( r = 0.25 \mathrm{~m} \). Thus, \( I = \frac{1}{2} \times 1.25 \times (0.25)^2 = 0.0390625 \mathrm{~kg} \cdot \mathrm{m}^2 \).
04

Calculate Rotational Kinetic Energy for the Smaller Cylinder

Use \( KE_{rot} = \frac{1}{2} I \omega^2 \) with \( I = 0.0390625 \mathrm{~kg} \cdot \mathrm{m}^2 \) and \( \omega = 235 \mathrm{~rad/s} \): \( KE_{rot} = \frac{1}{2} \times 0.0390625 \times (235)^2 = 1080.5 \mathrm{~J} \).
05

Determine Moment of Inertia for the Larger Cylinder

Substitute the given values for the larger cylinder into the inertia formula: \( m = 1.25 \mathrm{~kg} \) and \( r = 0.75 \mathrm{~m} \). Thus, \( I = \frac{1}{2} \times 1.25 \times (0.75)^2 = 0.3515625 \mathrm{~kg} \cdot \mathrm{m}^2 \).
06

Calculate Rotational Kinetic Energy for the Larger Cylinder

Use \( KE_{rot} = \frac{1}{2} I \omega^2 \) with \( I = 0.3515625 \mathrm{~kg} \cdot \mathrm{m}^2 \) and \( \omega = 235 \mathrm{~rad/s} \): \( KE_{rot} = \frac{1}{2} \times 0.3515625 \times (235)^2 = 9724.5 \mathrm{~J} \).

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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 a property that measures an object's resistance to changes in its rotational motion. It's essentially the rotational equivalent of mass in linear motion. For a solid cylinder rotating around its central axis, the formula to calculate the moment of inertia is straightforward:\[ I = \frac{1}{2} m r^2 \]Here:
  • \( I \) stands for the moment of inertia
  • \( m \) represents the mass of the cylinder
  • \( r \) is the radius of the cylinder
This equation tells us that the further the mass is from the axis of rotation (increased radius), the greater the moment of inertia. Hence, a larger cylinder with the same mass, but greater radius, will have a higher moment of inertia compared to a smaller cylinder. This principle helps us understand why larger cylinders take more energy to start and stop spinning.
Angular Velocity
Angular velocity describes how fast an object is rotating and is measured in radians per second (rad/s). It shows the rate of change of the angular position of an object and plays a pivotal role in rotational motion dynamics.The concept of angular velocity is crucial in solving problems involving rotational kinetic energy. For our problem, both cylinders are rotating at the same angular velocity \( \omega = 235 \mathrm{~rad/s} \). This uniform speed allows us to focus on variations due to their size differences.Angular velocity directly affects the rotational kinetic energy of an object through the equation:\[ KE_{rot} = \frac{1}{2} I \omega^2 \]A key takeaway is that even if two objects rotate at the same speed, differences in their size and mass distribution (moment of inertia) will impact the total energy associated with their rotation.
Solid Cylinder
A solid cylinder is a three-dimensional geometric shape with two congruent faces - circular bases, and a curved surface connecting these bases. It is a common shape for many practical objects like batteries, rollers, and barrels.When analyzing rotational motion, considering the characteristics of a solid cylinder becomes important. For a solid cylinder rotating around its central axis, the mass and radius are key to determining its moment of inertia. This specific structure affects how it behaves in rotational motion, which in turn influences calculations like rotational kinetic energy.In our example:
  • The smaller cylinder has a radius of \(0.25 \mathrm{~m}\) resulting in a smaller moment of inertia.
  • The larger cylinder has a radius of \(0.75 \mathrm{~m}\), making it three times larger in radius, and consequently it has a greater moment of inertia.
These physical attributes of a cylinder play a significant role in understanding and solving physical problems involving rotation.

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Most popular questions from this chapter

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0\), what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s}\) ? (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?

(a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} . $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

If an airplane propeller rotates at 2000 rev/min while the airplane flies at a speed of \(480 \mathrm{~km} / \mathrm{h}\) relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius \(1.5 \mathrm{~m}\), as seen by (a) the pilot and (b) an observer on the ground? The plane's velocity is parallel to the propeller's axis of rotation.

A high-wire walker always attempts to keep his center of mass over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of \(70.0 \mathrm{~kg}\) and a rotational inertia of \(15.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the wire. What is the magnitude of his angular acceleration about the wire if his com is \(5.0 \mathrm{~cm}\) to the right of the wire and (a) he carries no pole and (b) the \(14.0 \mathrm{~kg}\) pole he carries has its \(\operatorname{com} 10 \mathrm{~cm}\) to the left of the wire?

What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius \(3220 \mathrm{~km}\) at a speed of \(29000 \mathrm{~km} / \mathrm{h}\) ?
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