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Answer:
Recall that torque is the turning effectiveness of a force. ... Position the circle so that it can rotate freely about a horizontal axis through its ... The flywheel is a disk with a 0.150-m radius.
Answer:
Dynamics of Rotational Motion: Rotational Inertia
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Understand the relationship between force, mass and acceleration.
Study the turning effect of force.
Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration.
If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 1. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton’s second law of motion. There are, in fact, precise rotational analogs to both force and mass.
The given figure shows a bike tire being pulled by a hand with a force F backward indicated by a red horizontal arrow that produces an angular acceleration alpha indicated by a curved yellow arrow in counter-clockwise direction.
Figure 1. Force is required to spin the bike wheel. The greater the force, the greater the angular acceleration produced. The more massive the wheel, the smaller the angular acceleration. If you push on a spoke closer to the axle, the angular acceleration will be smaller.
To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force F on a point mass m that is at a distance r from a pivot point, as shown in Figure 2. Because the force is perpendicular to r, an acceleration\displaystyle a=\frac{F}{m}a=
m
F
is obtained in the direction of F. We can rearrange this equation such that F = ma and then look for ways to relate this expression to expressions for rotational quantities. We note that a = rα, and we substitute this expression into F = ma, yielding
F = mrα
Recall that torque is the turning effectiveness of a force. In this case, because F is perpendicular to r, torque is simply τ = Fr. So, if we multiply both sides of the equation above by r, we get torque on the left-hand side. That is,
rF = mr2α
or
τ = mr2α.
This last equation is the rotational analog of Newton’s second law (F = ma) where torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass (or inertia). The quantity mr2 is called the rotational inertia or moment of inertia of a point mass m a distance r from the center of rotation.
Explanation:
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