PHY101 2nd Assignment Solution

Q# 1 A compact disc stores lectures Physics (PHY101) in a coded pattern of tiny pits 10^-7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 20.0mm and 50.0mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.5m/s.

(a) What is the angular speed of the CD when the innermost part of the track is scanned the outermost part of the track?

(b) The maximum playing time of a CD is 75.0m. What would be the length of the track on such a maximum duration CD if it were stretched out in a straight line?

(c) What is the average angular acceleration of a maximum duration CD during its 75.0m playing time? Take the direction of rotation of the disc to be positive.

Ans 1. The problem statement, all variables and given/known data
Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits 10^-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s.
The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
2. Relevant equations
theta = (w_f + w_0)t
s = r*theta
w = angular velocity
3. The attempt at a solution
This is part 3 of the problem so I already have the angular velocity for the innermost and outermost of the disc--which I assume it is the initial and final angular velocity. Pls correct me if I'm wrong. I've used the eqns above to find theta and the length but I'm coming out with the wrong answer, 5.25km. I took the outer radius - inner radius = r. Can someone pls correct me? I don't know what I'm doing wrong. :frown:

Q # 2

The flywheel of a gasoline engine is required to give up 550J of kinetic energy while its angular velocity decreases from 790 rev/min to 730rev/min. What moment of inertia is required?

Ans

790rpm = 790 x 2pi / 60 rad/s; w1 = 82.73 rad/s
730rpm = 730 x 2pi / 60 rad/s; w2 = 76.45 rad/s
Kineic energy of rotation = Iw2/2 so
550 =(I x 82.73^ /2) - (I x 76.45^2)
1100 = I x (82.73^2 - 76.45^2)
I = 1100 / 999.65
= 1.10 kgm^2

Phy101 Assignment No. 1 solution

PHYSICS (PHY101)

TOTAL MARKS: 20

Due Date: 18/04/2011

Question No. 1

If you have an umbrella, suddenly you have to face a rain storm with a strong wind, what determines the best position in which to hold an umbrella? Marks = 3

Answer:

The raindrops have a vertical velocity relative to the ground & you have a horizontal velocity when moving relative to the ground & so the raindrops can penetrate "under" the umbrella & wet your legs.

Question # 2

Does Newton‘s second law hold true for an observer in a car as it speeds up, slows down or rounds a corner?

Answer: Marks = 5

Yes it holds true. An inertial reference frame is one in which Newton's First Law of Motion, the Law of Inertia, holds true. In such a frame, objects which have no force acting on them move in straight lines, with constant speed. Any deviation from such a motion is presumed to be due to a force whose magnitude and direction can be deduced by applying Newton's Second Law of Motion, the Force Law, to observations of that deviation.

An observer who is at rest is in an inertial reference frame, and observers who are not at rest, but are moving with uniform (unchanging, straight-line) motion are also in inertial reference frames. Any acceleration, whether a change in speed, a change of direction (such as, in the example above, our motion around the Earth's axis of rotation), or both, puts observers who share in that acceleration in an ac The observer inside the car, however, tends to use the car as his reference frame, just the same as if it were still at rest, and since it is accelerating, will perceive things differently from the observer who really is at rest. celerated reference frame.

Question No. 3

A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, will the distance between them increase, decrease, or remain the same? Prove your answer. Mark 5Answer:By supposing above scenario we ll apply this equation Change in velocity= change in position /travel time According to this equation with the fall successively downward the average velocity will increase and hence the distance between them will also increase

Question # 4

The fastest measured pitched baseball left the pitcher’s hand at a speed of 55.0m/s. if the pitcher was in contact with the ball over a distance of 2.0m and produce constant acceleration,

(a) what acceleration did he give the ball, and

(b) How much time did it take him to pitch it?

Answer: Marks = 7

(a) What acceleration did he give the ball

2as= vf2-vi22

(a)(2)= (55)^2-(0)

4(a) = 3 025

(a)=3025/4

a=756.25 m/s^2

(b) How much time did it take him to pitch it

t = v - u / a

= 55-0/756.25

=55 sec