Practice Problems: Uniform Circular Motion Solutions

1. (moderate) A racecar, moving at a constant tangential speed of 60 m/s, takes one lap around a circular track in 50 seconds. Determine the magnitude of the acceleration of the car.
a = v²/r
T = 2πr/v.....r = Tv/2π
combine...a = v²/(Tv/2π)= v/(T/2π)
a = (60)/(50/6.28) = 7.5 m/s²

2. (moderate) An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s². If the radius of the motion is 0.02 m, what is the frequency of the motion?
a = v²/r
13 = v²/0.02
v = 0.51 m/s
v = 2πr/T
0.51 = 6.28(0.02)/T
T = 0.25 s
f = 1/T
f = 1/(0.25) = 4 Hz

3. (easy) Find the centripetal acceleration for an object on the surface of a planet (at the equator) with the following characteristics: radius = r = 4x10⁶ m and 1 day = 100000 seconds.
a = v²/r = (2πr/T)²/r = 4π²r/T²
a = 4(3.14)²(4x10⁶)/(100000)² = 0.016 m/s²

4.(moderate) For an object in uniform circular motion rank the changes listed
below regarding the effect each would produce on the magnitude of the
centripetal acceleration of the object? Assume all other parameters stay
constant except that noted in the description of the change.
Change A: The speed of the object doubles
Change B: The radius of the motion triples
Change C: The mass of the object triples
Change D: The radius of the motion becomes half as big
Change E: The mass of the object becomes half as big
Change F: The speed of the object becomes half as big
Since a = v²/r, the mass doesn’t affect the acceleration magnitude
as long as the speed and the radius stays the same. Of course, the object would
have to be forced differently to keep the speed the same if the mass changes.
We will discuss this concept when we cover Newton’s 2^nd Law.
Change A: The acceleration quadruples
Change B: The acceleration becomes 1/3 as big
Change C: The acceleration doesn’t change
Change D: The acceleration doubles
Change E: The acceleration doesn’t change
Change F: The acceleration becomes 1/4 as big
Change F < Change B < Change C = Change E < Change D < Change A

5. (moderate) Outline a procedure by which the centripetal acceleration of a car moving with constant speed on a circular racetrack could be determined. List the standard lab apparatuses needed to make the measurements and the calculations a student can make with the measurements to determine the acceleration. Additionally, determine the range of error introduced into the final answer if each measurement could have an error of 10%. This type of question is very important for your test preparation. Please take your time and answer it completely.
Measurement devices needed:
A long measuring tape
A stopwatch
Step 1: Use the measuring tape to determine the radius (r) of the path of the
car on the circular racetrack. Model the car as a point particle. Use the
geometric center of the car as the location of this point.
Step 2: Use the stopwatch to determine the time (T) needed for the car to move
once around the track.
Calculations:
speed = v = 2πr/T
Determine the acceleration: a = v²/r = (2πr/T)²/r = 4π²r/T²

Since the each measurement has a possible error of 10%, the worst case scenario
would be if the radius measurement was 10% too high and the time measurement
was 10% too low.
Example:
If the actual radius is 10.0 m and the radius measurement is 11.0 m
If the actual time is 10.0 s and the time measurement is 9.0 s
Actual acceleration = a = 4π²(10.0)/(10.0)² = 0.40π² m/s²
Measured acceleration = a_m = 4π²(11.0)/(9.0)² = 0.54π² m/s²
% Error = [(Measured – Actual)/Actual]100% %
Error = [(0.54π² – 0.40π²)/0.40π²]100% = +35 %
These measurements would result in an answer that is 35% too high. This is the
largest expected error that is too high.

However, one can also determine an answer that is too low. This occurs at its maximum level when the radius is 10% too low and the time measurement is 10% too high:
Example:
If the actual radius is 10.0 m and the radius measurement is 9.0 m
If the actual time is 10.0 s and the time measurement is 11.0 s
Actual acceleration = a = 4π²(10.0)/(10.0)² = 0.40π² m/s²
Measured acceleration = a_m = 4π²(9.0)/(11.0)² = 0.30π² m/s²
% Error = [(Measured – Actual)/Actual]100%
% Error = [(0.30π² – 0.40π²)/0.40π²]100% = -25 %

When the other extremes occur, the % error is in between the calculations shown
above.
Example:
If the actual radius is 10.0 m and the radius measurement is 11.0 m
If the actual time is 10.0 s and the time measurement is 11.0 s
Actual acceleration = a = 4π²(10.0)/(10.0)² = 0.40π² m/s²
Measured acceleration = a_m = 4π²(11.0)/(11.0)² = 0.36π² m/s²
% Error = [(Measured – Actual)/Actual]100%
% Error = [(0.36π² – 0.40π²)/0.40 ²]100% = -10 %
Example:
If the actual radius is 10.0 m and the radius measurement is 9.0 m
If the actual time is 10.0 s and the time measurement is 9.0 s
Actual acceleration = a = 4π²(10.0)/(10.0)² = 0.40π² m/s²
Measured acceleration = a_m = 4π²(11.0)/(11.0)² = 0.44π² m/s²
% Error = [(Measured – Actual)/Actual]100%
% Error = [(0.44π² – 0.40π²)/0.40π²]100% = +10 %

When the measurements are not at the extremes, the % error again will fall in
between its maximum positive and maximum negative values.
Example:
If the actual radius is 10.0 m and the radius measurement is 10.5 m
If the actual time is 10.0 s and the time measurement is 9.5 s
Actual acceleration = a = 4π²(10.0)/(10.0)² = 0.40π² m/s²
Measured acceleration = a_m = 4π²(10.5)/(9.5)² = 0.47π² m/s²
% Error = [(Measured – Actual)/Actual]100%
% Error = [(0.47π² – 0.40π²)/0.40π ²]100% = +18 %

6. (moderate) A particle is moving at a constant speed in a circular trajectory centered on the origin of an xy coordinate system. At one point (x = 4 m, y = 0 m) the particle has a velocity of -5.0j m/s. Determine the velocity and acceleration when the particle is at:
a. x = 0, y = -4 m
The velocity is always tangent to the trajectory.
v = -5i m/s (because the circular motion is clockwise on the axis system)
a = v²/r = 5²/4 = 6.25 m/s²
a = 6.25j m/s²(always directed toward the center)
b. x = -4 m, y = 0
v = 5j m/s
The acceleration magnitude is constant, so we still have a = 6.25 m/s²
a = 6.25i m/s² (always directed toward the center)
c. x = 0, y = 4 m
v = 5i m/s
a = -6.25j m/s² 
d. x = -2.83, y = 2.83 m
The velocity will still have a magnitude of 5 m/s, and will still be tangent to the trajectory. At the point referred to, the angle of the tangent is 45°.
v = 5m/s(cos45i + sin45j)
v = 3.54i +3.54j m/s
The acceleration will still have the same magnitude, and again, must point toward the center. In this case, at a 315° angle.
a = 6.25 m/s²(cos315i + sin315j)
a = 4.42i - 4.42j m/s²

7. (moderate) A stunt pilot executes a uniform speed circular path in an airplane. The initial velocity (in m/s) of the plane is given by v_o = 2500i + 3000j. One minute later the velocity of the plane is v = -2500i - 3000j. Find the magnitude of the acceleration.
First find the speed:
v = (2500² + 3000²)^½ = 3905 m/s
For uniform circular motion:
a = v²/r and T = 2πr/v
Combine to show that a = 2πv/T
Since T is the period of the motion, and the given data report that it takes one minute to reverse the velocity (the components have reversed), the period is 2 minutes (120 s).
a = 2π(3905)/120
a = 204 m/s²

8. (moderate) This problem is not refering to an object in uniform circular motion, but it deals with motion in two dimensions. The velocity (in m/s) of a particle moving in the x-y plane is given by:
v = ( 6.0t – 4.0t²)i + 8.0 j
a. What is the acceleration when t = 3.0 seconds?
a = dv/dt
a = (6.0 - 8.0t)i
a = (6.0 - 24.0)i = -18.0i m/s²
b. When, if ever, is the acceleration zero?
a = 0 = 6.0 - 8.0t
6.0 = 8.0t
t = 6.0/8.0 = 0.75 s
c. When, if ever, is the speed zero?
The speed can never be zero, because there is always a component of the velocity in the j direction.