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Chapter 10: Q4Q (page 286)

Figure 10-22 bis a graph of the angular position of the rotating disk of Fig.10-22 a. Is the angular velocity of the disk positive, negative, or zero at (a)t=1s, (b)t=2s, and (c)t=3s? (d) Is the angular acceleration positive or negative?

Short Answer

Expert verified
  1. Angular velocities of the disk, at pointt=1 sis positive.
  2. Angular velocities of the disk, at pointt=2 sis zero.
  3. Angular velocities of the disk, at pointt=3sis negative.
  4. The angular acceleration is negative.

Step by step solution

01

Step 1: Given data

Graph of angular position vs time.

02

Determining the concept

Find the sign of the slope of the graph which represents the sign of angular velocity. Conclude about the angular acceleration from the nature of the graph.

03

(a) Determining if the angular velocity of the disk is positive, negative, or zero at  t= 1 s

The given graph is an angular position vs time graph, so the slope of the graph is angular velocity.

At point t=1, it is positive because the slope is positive.

04

(b) Determining if the angular velocity of the disk is positive, negative, or zero at t= 2 s 

At point t=2, it is zero because the slope is zero.

05

(c) Determining if the angular velocity of the disk is positive, negative, or zero at t= 3 s

At point t=3, it is negative, because slope is negative.

06

(d) Determining if the angular acceleration is positive or negative

The given fact that the original function is concave down, its second derivative must be negative.

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Most popular questions from this chapter

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.0s, it rotates 25rad. During that time, what are the magnitudes of

(a) the angular acceleration and

(b) the average angular velocity?

(c) What is the instantaneous angular velocity of the disk at the end of the 5.0s?

(d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next 5.0s?

In a judo foot-sweep move, you sweep your opponent’s left foot out from under him while pulling on his gi (uniform) toward that side. As a result, your opponent rotates around his right foot and onto the mat. Figure 10−44shows a simplified diagram of your opponent as you face him, with his left foot swept out. The rotational axis is through point O. The gravitational force F→g on him effectively acts at his center of mass, which is a horizontal distanced=28 cm from point O. His mass is 70Kg, and his rotational inertia about point O is 65 kg.m2.What is the magnitude of his initial angular acceleration about point O if your pull F→a on his gi is (a) negligible and (b) horizontal with a magnitude of 300Nand applied at height h=1.4 m ?

A bicyclist of mass 70kg puts all his mass on each downward moving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be0.40m , and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

A disk, with a radius of0.25 m , is to be rotated like a merrygo-round through , starting from rest, gaining angular speed at the constant rate α1through the first400 rad and then losing angular speed at the constant rate-α1 until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed400 ms2 .

(a) What is the least time required for the rotation?

(b) What is the corresponding value ofα1 ?

(a) If R=12 cm ,M=400 g , andm=50 g in Fig.10−19 , find the speed of the block after it has descended50 cm starting from rest. Solve the problem using energy conservation principles.

(b) Repeat (a) with R=5.0 cm.
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