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 / s²)(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 / s² x 5s) = 1.25 rad / s

c) Angular acceleration: remains constant at 0.10 rad / s²

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 t² as follows:

q(3s) = 0 rad + (16.7 rad / s)(3s) + 1/2(-5.5p rad / s²)(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.