Q1: How large are these
angles in radians? A half rotation, a full rotation, 5 rotations,
45 degrees, 90 degrees, 360degrees.
Half rotation (180°): 180° x 2p
rad / 360° = p rad
Full rotation (360°): 360° x 2p
rad / 360° = 2p rad
5 rotations (5x360°): 1800° x 2p
rad / 360° = 10p rad
45 degrees: 45° x 2p rad / 360°
= p rad / 4
90 degrees: 90° x 2p rad / 360°
= p rad / 2
360 degrees: 360° x 2p rad /
360° = 2p rad
Q2: A
car engine rotates at a rate of 4500 rpm (revolutions per minute).
What is the angular velocity of the engine in radians / sec?
Q3: A
man of height 180 cm subtends an angle of .0036 radians. How far
away is the man?
To solve this problem we must employ the
equation: q = s / r
r = s / q = 1.8m
/ 0.0036 rad = 500 m
The angular position of a merry-go-round
after a switch is thrown is specified as:
Q4: What
are the inital angular position, angular velocity and angular acceleration
of the merry-go-round before the switch is thrown? (Hint: compare
to the linear kinematic equations.)
Q5: What
are the angular position, velocity and acceleration of the merry-go-round
at t = 5 seconds.
a) Angular position: q
(5s) = [1.23 rad + (0.75 rad / s)(5s) + 1/2(0.10 rad / s^{2})(5s)^{2}]
q
(5s) = 1.23 rad + 3.75 rad + 1.25 rad = 6.23 rad
b) Angular velocity: w
= w_{o} + a t, which is
related to the linear equation v = v_{o} + at
w = 0.75
rad / s + (0.10 rad / s^{2}
x 5s) = 1.25 rad / s
c) Angular acceleration: remains constant
at 0.10 rad / s^{2}
An elecrical motor running at 1000 rpm is turned
off, and its angular velocity decreases uniformly to 500 rpm over
three seconds.
Q6:
What is the angular acceleration of the motor?
To solve this problem, we can rearrange the equation w
= w_{o} + at to solve
for a. However, we must first convert
the angular velocities into rad/s as follows:
Now, rearranging w = w_{o}
+ a t to solve for a,
we get: a = (w -
w_{o}) / t
a = (16.7p
rad / s - 33.3p rad / s) / 3s
= -5.5p rad / s^{2 }^{(The
negative sign indicates that the motor is slowing down.}
Q7: If
the motor continues to decelerate uniformly from 500 to 0 rpm,
how many revolutions does the motor turn through before coming
to a rest?
To solve this problem, we must employ
the equation q(t) = q_{o} + wt
+ 1/2a t^{2} as follows:
q(3s)
= 0 rad + (16.7 rad / s)(3s) + 1/2(-5.5p rad
/ s^{2})(3s)^{2}
q(3s) = 50.1 rad - 24.75p rad
= 25.35 p rad = 12.7 revolutions
Q8: If
the motor continues to uniformly coast to a halt, how much time
is required to come to a complete stop?
This question can be answered intuitively. If we know that
the motor is decelerating uniformly, and it took 3 seconds to
go from 1000 rpm to 500 rpm, then we know that it will take an
additional 3 seconds to reach a halt. |