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Chapter 9: Problem 18

A CD starts from rest and speeds up to the operating angular frequency of the CD player. Compare the angular velocity and acceleration of a point on the edge of the CD to those of a point halfway between the center and the edge of the CD. Do the same for the linear velocity and acceleration.

Short Answer

Expert verified
Answer: The angular velocity and acceleration are the same for both points since they are on the same rotating CD. However, the linear velocity and acceleration of the point on the edge are twice the linear velocity and acceleration of the point halfway between the center and the edge.

Step by step solution

01

Identify knowns and unknowns

In this problem, we know the CD starts from rest and accelerates to its operating angular frequency (ω). We are asked to find the angular and linear values of velocity and acceleration for two points on the CD. Let's define two points: - Point A: On the edge of the CD - Point B: Halfway between the center and the edge of the CD As both points are on the same CD, they share the same angular characteristics, i.e. ω and α. We just need to find the linear variables, v and a, for each point.
02

Angular Variables

Since both points are on the same rotating CD, they will have the same angular velocity (ω) and angular acceleration (α). Therefore: ω_A = ω_B = ω α_A = α_B = α
03

Linear Variables

For the linear variables, we can use the following relationships: v = rω a = rα To find the exact values, we need the radius of the CD and the distance between the center and point B. However, as our task is to compare the two points, we can analyze them using ratios. Let R be the radius of the CD. Then, we have: v_A = Rω v_B = (R/2)ω Now, we can see that the linear velocity of point A is twice the linear velocity of point B: v_A = 2v_B Next, let's analyze the linear acceleration: a_A = Rα a_B = (R/2)α Similarly, the linear acceleration of point A is twice the linear acceleration of point B: a_A = 2a_B
04

Conclusion

In this exercise, we have compared the angular and linear variables of velocity and acceleration of two points on a CD. We have found that, since both points are on the same rotating CD, they have the same angular velocity (ω) and angular acceleration (α): ω_A = ω_B = ω α_A = α_B = α However, the linear velocity and acceleration are different due to their varying distances from the center: v_A = 2v_B a_A = 2a_B Thus, the point on the edge of the CD (point A) experiences twice the linear velocity and acceleration as the point halfway between the center and the edge (point B).

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Most popular questions from this chapter

A ring is fitted loosely (with no friction) around a long, smooth rod of length \(L=0.50 \mathrm{~m} .\) The rod is fixed at one end, and the other end is spun in a horizontal circle at a constant angular velocity of \(\omega=4.0 \mathrm{rad} / \mathrm{s} .\) The ring has zero radial velocity at its initial position, a distance of \(r_{0}=0.30 \mathrm{~m}\) from the fixed end. Determine the radial velocity of the ring as it reaches the moving end of the rod.

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