Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
Chapter 9, Problem 1P

(a)

To determine

The point moves a greater distance in a given time, if the disk rotates with increasing angular velocity.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The particle on the rim covers a greater linear distance as compared to the particle situated half way between the rim and the axis of rotation.

Explanation of Solution

Given:

A disk rotating with increasing angular velocity about an axis passing through its center and perpendicular to its plane.

Distance of the particle 1 (on the rim) from the axis of rotation

  r1=r

Distance of particle 2 (half way between the axis of rotation and the rim) from the axis of rotation

  r2=r2

Here, the radius of the disk is r

Formula used:

The angular acceleration α of the disk is given by

  α=dωdt ……(1)

Here, ω is the angular velocity of the disk.

The angular acceleration of a point on the rotating disk is related to the linear acceleration a as,

  a=rα……(2)

Here, r is the distance of the point from the axis of rotation.

The linear distance Δx travelled by a particle in a given time interval t is given by

  Δx=v0t+12at2……(3)

Calculations:

The disk of radius r rotates with increasing angular velocity about an axis passing through its center and perpendicular to the plane. Point 1 is located at its rim and point 2 is located at a distance r2

from the axis of rotation. This is shown in the figure 1 below:

  Physics for Scientists and Engineers, Chapter 9, Problem 1P , additional homework tip  1

  Figure 1

If the disk rotates with an angular acceleration α , both points 1 and 2 have the same angular acceleration and they would have the same angular displacement during any time interval considered.

Calculate the linear accelerations of the particles located at point 1 and 2 using the equation (2).

  a1=r1αa2=r2α……(4)

Substitute r1=r and r2=r2 in equation (4).

  a1=r1α=rα

  a2=r2α=r2α

Therefore, from the above equations,

  a2=12a1……(5)

If the disk is assumed to start from rest, both particles would start with their initial velocities v0 which would be equal to zero.

Use equation (3) to calculate the distance travelled by the two points.

  Δx1=12a1t2Δx2=12a2t2

From equation (5),

  Δx1=2Δx2

Conclusion:

Therefore, the point on the rim travels a greater distance when compared to the point located halfway between the rim and the axis of rotation.

(b)

To determine

The point that turns through a greater angle.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

Both the points turn through the same angle.

Explanation of Solution

Formula used:

The angular displacement Δθ of a particle in a time interval t is given by

  Δθ=ω0t+12αt2…..(6)

Here ω0 is the initial angular velocity of the particle.

Calculation:

The points 1 and 2 located at points A and B on the disc rotating with an angular acceleration α in the clockwise direction move to positions A' and B' in the time t as shown in Figure 2.

  Physics for Scientists and Engineers, Chapter 9, Problem 1P , additional homework tip  2

  Figure 2

At any instant of time, both particles have the same instantaneous angular velocity and angular acceleration. As it can be seen from Figure 2, both particles describe the same angle at an instant of time.

Conclusion:

Thus, the particle located at the rim and the particle located half way between the rim and the axis of rotation turn through the same angle.

(c)

To determine

The point which travels with greater speed.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

The point on the rim travels with greater speed.

Explanation of Solution

Formula used:

The instantaneous speed v of a point located at a distance r from the axis of rotation is given by

  v=rω……(7)

Here, ω is the instantaneous angular velocity.

Calculation:

The disk moves with increasing angular velocity. But, both points at any instant would have the same instantaneous angular velocity, since they turn through the same angle in a given interval of time.

Hence, it can be inferred from equation (7):

  vr

Since the point on the rim has the greater value of r when compared to the point located inside, it would have a greater speed.

Conclusion:

Thus, the point on the rim would have a greater speed when compared to the point located half way between the rim and the axis of rotation.

(d)

To determine

The point which has the greater angular speed.

(d)

Expert Solution
Check Mark

Answer to Problem 1P

Both the particles have the same angular speed.

Explanation of Solution

Formula used:

The angular velocity of a particle is given by

  ω=dθdt……(8)

Calculation:

From Figure 2, it is seen that at any instant of time, both particles 1 and 2 cover the same angles. Hence, the rate of change of their angular displacement dθdt will be the same. Thus, it follows that both the points will have the same angular velocity at any instant of time, according to equation (8).

Conclusion:

Thus, the particle on the rim and the particle located halfway between the rim and the axis of rotation have the same angular velocity.

(e)

To determine

The point which has the greater tangential acceleration.

(e)

Expert Solution
Check Mark

Answer to Problem 1P

The point on the rim has a greater tangential acceleration when compared to the point located midway between the rim and the axis of rotation:

Explanation of Solution

Formula used:

The tangential acceleration at of a point on a rotating disk located at a distance r from the axis of rotation is given by

  at=rα……(9) Here, α is the angular acceleration of the disk.

Calculation:

If the disk rotates with a varying angular velocity, it has angular acceleration. Assuming that the angular acceleration of the disk remains constant, from equation (9) it can be inferred that

  atr

The point on the rim has the greater value of r when compared to the point located halfway between the rim and the axis of rotation.

Therefore, the point on the rim has a greater tangential acceleration when compared to any point located inside the rim. This is also proved by the fact that the point on the rim gains a larger tangential velocity when compared to any inner point.

Conclusion:

Thus, the point on the rim has a greater tangential acceleration when compared to the point located midway between the rim and the axis of rotation.

(f)

To determine

The point which has a greater angular acceleration.

(f)

Expert Solution
Check Mark

Answer to Problem 1P

Both particles have the same angular acceleration.

Explanation of Solution

Formula used:

The angular acceleration α of the disk is given by

  α=dωdt……(1)

Calculation:

It has been proved in (d) that at any instant of time, both the particles have the same angular velocity. Therefore, in an interval of time dt , the change in their angular velocities dω will also be equal.

Hence, from equation (1), it can be proved that at a given instant of time, both the points will have the same angular accelerations.

Conclusion:

Thus, both particles are found to have the same angular acceleration.

(g)

To determine

The point which has the greater centripetal acceleration.

(g)

Expert Solution
Check Mark

Answer to Problem 1P

The point on the rim has a greater centripetal acceleration.

Explanation of Solution

Formula used:

The centripetal acceleration of a point located at a distance r from the axis of rotation is given by

  ar=rω2……(10)

Calculation:

It has been established in part (d) that at any instant of time, the point on the rim and the point located halfway between the rim and the axis of rotation have the same angular velocity.

Therefore, from equation (10), it can be inferred that

  arr

The point on the rim has the greater value of r when compared to the point located in between the rim and the axis of rotation. Therefore, the centripetal acceleration of the point on the rim would be greater than the centripetal acceleration of the point located in between the rim and the axis of rotation.

Conclusion:

Thus, the point on the rim has a greater centripetal acceleration.

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Chapter 9 Solutions

Physics for Scientists and Engineers

Ch. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Prob. 13PCh. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 26PCh. 9 - Prob. 27PCh. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - Prob. 31PCh. 9 - Prob. 32PCh. 9 - Prob. 33PCh. 9 - Prob. 34PCh. 9 - Prob. 35PCh. 9 - Prob. 36PCh. 9 - Prob. 37PCh. 9 - Prob. 38PCh. 9 - Prob. 39PCh. 9 - Prob. 40PCh. 9 - Prob. 41PCh. 9 - Prob. 42PCh. 9 - Prob. 43PCh. 9 - Prob. 44PCh. 9 - Prob. 45PCh. 9 - Prob. 46PCh. 9 - Prob. 47PCh. 9 - Prob. 48PCh. 9 - Prob. 49PCh. 9 - Prob. 50PCh. 9 - Prob. 51PCh. 9 - Prob. 52PCh. 9 - Prob. 53PCh. 9 - Prob. 54PCh. 9 - Prob. 55PCh. 9 - Prob. 56PCh. 9 - Prob. 57PCh. 9 - Prob. 58PCh. 9 - Prob. 59PCh. 9 - Prob. 60PCh. 9 - Prob. 61PCh. 9 - Prob. 62PCh. 9 - Prob. 63PCh. 9 - Prob. 64PCh. 9 - Prob. 65PCh. 9 - Prob. 66PCh. 9 - Prob. 67PCh. 9 - Prob. 68PCh. 9 - Prob. 69PCh. 9 - Prob. 70PCh. 9 - Prob. 71PCh. 9 - Prob. 72PCh. 9 - Prob. 73PCh. 9 - Prob. 74PCh. 9 - Prob. 75PCh. 9 - Prob. 76PCh. 9 - Prob. 77PCh. 9 - Prob. 78PCh. 9 - Prob. 79PCh. 9 - Prob. 80PCh. 9 - Prob. 81PCh. 9 - Prob. 82PCh. 9 - Prob. 83PCh. 9 - Prob. 84PCh. 9 - Prob. 85PCh. 9 - Prob. 86PCh. 9 - Prob. 87PCh. 9 - Prob. 88PCh. 9 - Prob. 89PCh. 9 - Prob. 90PCh. 9 - Prob. 91PCh. 9 - Prob. 92PCh. 9 - Prob. 93PCh. 9 - Prob. 94PCh. 9 - Prob. 95PCh. 9 - Prob. 96PCh. 9 - Prob. 97PCh. 9 - Prob. 98PCh. 9 - Prob. 99PCh. 9 - Prob. 100PCh. 9 - Prob. 101PCh. 9 - Prob. 102PCh. 9 - Prob. 103PCh. 9 - Prob. 104PCh. 9 - Prob. 105PCh. 9 - Prob. 106PCh. 9 - Prob. 107PCh. 9 - Prob. 108PCh. 9 - Prob. 109PCh. 9 - Prob. 110PCh. 9 - Prob. 111PCh. 9 - Prob. 112PCh. 9 - Prob. 113PCh. 9 - Prob. 114PCh. 9 - Prob. 115PCh. 9 - Prob. 116PCh. 9 - Prob. 117PCh. 9 - Prob. 118PCh. 9 - Prob. 119PCh. 9 - Prob. 120PCh. 9 - Prob. 121PCh. 9 - Prob. 122PCh. 9 - Prob. 123PCh. 9 - Prob. 124PCh. 9 - Prob. 126PCh. 9 - Prob. 127PCh. 9 - Prob. 128PCh. 9 - Prob. 129P
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