To find the total angular displacement during the playing time of the compact disc in part (B) of Example 10.2, the disc was modeled as a rigid object under constant angular acceleration . In reality, the angular acceleration of a disc is not constant. In this problem, let us explore the actual time dependence of the angular acceleration. (a) Assume the track on the disc is a spiral such that adjacent loops of the track are separated by a small distance h . Slum that the radius r of a given portion of the track is given by r = r i + h θ 2 π where r i is the radius of the innermost portion of the track and θ is the angle through which the disc turns to arrive at the location of the track of radius r . (b) Show that the rate of change of the angle θ is given by d θ d t = v r i + ( h θ / 2 π ) where v is the constant speed with which the disc surface passes the laser. (c) From the result in part (b), use integration to find an expression for the angle θ as a function of time. (d) From the result in part (c), use differentiation to find the angular acceleration of the disc as a function of time.
To find the total angular displacement during the playing time of the compact disc in part (B) of Example 10.2, the disc was modeled as a rigid object under constant angular acceleration . In reality, the angular acceleration of a disc is not constant. In this problem, let us explore the actual time dependence of the angular acceleration. (a) Assume the track on the disc is a spiral such that adjacent loops of the track are separated by a small distance h . Slum that the radius r of a given portion of the track is given by r = r i + h θ 2 π where r i is the radius of the innermost portion of the track and θ is the angle through which the disc turns to arrive at the location of the track of radius r . (b) Show that the rate of change of the angle θ is given by d θ d t = v r i + ( h θ / 2 π ) where v is the constant speed with which the disc surface passes the laser. (c) From the result in part (b), use integration to find an expression for the angle θ as a function of time. (d) From the result in part (c), use differentiation to find the angular acceleration of the disc as a function of time.
Solution Summary: The author explains that the radius of a given portion of the track is given by r=r_i+htheta 2pi
To find the total angular displacement during the playing time of the compact disc in part (B) of Example 10.2, the disc was modeled as a rigid object under constant angular acceleration. In reality, the angular acceleration of a disc is not constant. In this problem, let us explore the actual time dependence of the angular acceleration. (a) Assume the track on the disc is a spiral such that adjacent loops of the track are separated by a small distance h. Slum that the radius r of a given portion of the track is given by
r
=
r
i
+
h
θ
2
π
where ri is the radius of the innermost portion of the track and θ is the angle through which the disc turns to arrive at the location of the track of radius r. (b) Show that the rate of change of the angle θ is given by
d
θ
d
t
=
v
r
i
+
(
h
θ
/
2
π
)
where v is the constant speed with which the disc surface passes the laser. (c) From the result in part (b), use integration to find an expression for the angle θ as a function of time. (d) From the result in part (c), use differentiation to find the angular acceleration of the disc as a function of time.
Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .
The de-excitation of a state occurs by competing emission and relaxation processes. If the relaxation mechanisms are very effective:a) the emission of radiation is largeb) the emission of radiation is smallc) the emission occurs at a shorter wavelengthd) the de-excitation occurs only by emission processes
m
C
A block of mass m slides down a ramp of height hand
collides with an identical block that is initially at rest.
The two blocks stick together and travel around a loop of
radius R without losing contact with the track. Point A is
at the top of the loop, point B is at the end of a horizon-
tal diameter, and point C is at the bottom of the loop, as
shown in the figure above. Assume that friction between
the track and blocks is negligible.
(a) The dots below represent the two connected
blocks at points A, B, and C. Draw free-body dia-
grams showing and labeling the forces (not com
ponents) exerted on the blocks at each position.
Draw the relative lengths of all vectors to reflect
the relative magnitude of the forces.
Point A
Point B
Point C
(b) For each of the following, derive an expression in
terms of m, h, R, and fundamental constants.
i. The speed of moving block at the bottom of
the ramp, just before it contacts the stationary
block
ii. The speed of the two blocks immediately…
The velocity of an elevator is given by the graph shown.
Assume the positive direction is upward.
Velocity (m/s)
3.0
2.5
2.0
1.5
1.0
0.5
0
0
5.0
10
15
20
25
Time (s)
(a) Briefly describe the motion of the elevator.
Justify your description with reference to the
graph.
(b) Assume the elevator starts from an initial position
of y = 0 at t=0. Deriving any numerical values
you
need from the graph:
i. Write an equation for the position as a
function of time for the elevator from
t=0 to t = 3.0 seconds.
ii. Write an equation for the position as a
function of time for the elevator from t = 3.0
seconds to t = 19 seconds.
(c) A student of weight mg gets on the elevator
and rides the elevator during the time interval
shown in the graph. Consider the force of con-
tact, F, between the floor and the student. How
Justify your answer with reference to the graph
does F compare to mg at the following times?
and your equations above.
i. = 1.0 s
ii. = 10.0 s
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