Rotational Rates

  This problem set deals with rotational rates.  
1. A wheel is 2 m in diameter. When it turns at 200 RPM, find the linear speed of a point on its circumference in m/s.

Let's look at a wheel turning at one revolution per minute. The rotational speed is 1 RPM for the entire wheel, but the linear speeds of different parts of the wheel depend on their distances (radii) from the center of the wheel. A point on the circumference of the wheel has to travel the entire circumference (which is equal to 2pr) in the one minute.

So, in order to find the linear speed of a point on a rotating object when we know the rotational speed, we first need to find the angle that the point traveled through in the time interval using the rotational speed equation, w = q/t. When the angle is in radians we can use the definition of the radian (q = s/r) to find the arc length, s, that the point travelled through in the time interval. This is the distance, d, in the speed equation (v = d/t).
 
Knowns Unknowns Equations/Remarks
D = 2 m
w = 200 RPM (rev/min)
d = s
1 rev = 2p rad
1 min = 60 s 		v = ? in m/s v = d/t
q = s/r
w = q/t
r = D/2
r = D/2
r = (2 m)/2 = 1 m

w = q/t
q = wt

q = s/r
s/r = wt
s = wrt
d = s = wrt

v = d/t
v = wrt/t
v = wr
v = (200 rev/min)(1 m)(2p rad/1 rev)(1 min/60 s)
v = 20.9 m/s

Let's start off easy and find the radius from the diameter.

Then rearrange the rotational speed equation to solve for the angle. The angle is also equal to the arc length divided by the radius (q = s/r). Substitute this in for the angle in the rearranged rotational speed equation. Now rearrange this equation to solve for the arc length. The arc length is the distance that the point travels through each time interval.

Substitute in for the distance. The t's divide to one, leaving the linear speed being equal to the rotational speed times the radius. Now substitute in the known values. Convert the rpm's to radians and the minutes to seconds. The radians are dimensionless units and are dropped in the final answer.
2. A shaft rotates through 500 revolutions in 25 seconds. Find its rotational speed in:

A. RPM
B. radians per second

  
Knowns Unknowns Equations/Remarks
q = 500 rev
t = 25 s
1 min = 60 s
1 rev = 2p rad 		A) w = ? in RPM

B) w = ? in rad/s

 w = q/t
A) w = q/t
    w = [(500 rev)/(25 s)](60 s/1 min)
    w = 1200 RPM

B) w = (1200 rev/min)(2p rad/1 rev)(1 min/60 s)
    w = 126 rad/s

A) Substitute in the known values. Convert seconds to minutes. Calculate the answer.

B) We can either convert the 1200 RPM to rad/s, or we can go back to the original equation and find the answer in rad/s instead of RPM. In this case, I converted the RPM to rad/s.
3. A wheel that starts from rest has an angular speed of 20 rad/s after being uniformly accelerated for 10 seconds. Find:

A. The average rotational speed
B. The total angle through which it turned in those 10 seconds.

This problem is similar to the average speed problem in the previous section. The average rotational speed is one-half the sum of the initial and final rotational speeds, wavg = (½)(wf + wi). The angle can then be found using the rotational speed equation, w = q/t.

 
Knowns Unknowns Equations/Remarks
wi = 0 rad/s
wf = 20 rad/s
t = 10 s		 A) wavg = ?

B) q = ?

wavg = (½)(wf + wi)

w = q/t

A) wavg = (½)(wf + wi)
    wavg = (1/2)(20 rad/s  + 0 rad/s)
    wavg = 10 rad/s

B) w = q/t
    q = wt
    q = (10 rad/s)(10 s)
    q = 100 radians

A) Substitute in the known values and calculate the average rotational speed.

B) Rearrange the rotational speed equation to solve for the angle. Substitute in the average rotational speed and the time. Calculate the answer.
4. A flywheel spinning clockwise completes 180 revolutions in 10 minutes. Find:

A. Its angular speed in RPM and rad/s
B. Its angular velocity

  
Knowns Unknowns Equations/Remarks
q = 180 rev
t = 10 min
1 min = 60 s
1 rev = 2p rad 		A) w = ? in RPM and rad/s

B) w = ? (angular velocity)

 w = q/t
A) w = q/t
    w = (180 rev)/(10 min) = 18 RPM

    w = (18 rev/min)(2p rad/1 rev)(1 min/60 s)
    w = 1.88 rad/s

B) The angular velocity would be 18 RPM CW and 1.88 rad/s CW (CW stands for clockwise).

 A) Substitute in the known values in the rotational speed equation and calculate. Convert RPM's to rad/s.

B) The angular or rotational speed is a scalar value, meaning that the direction is not included. The angular or rotational velocity is a vector, meaning that it has a magnitude (speed) and direction. In this case, the direction is clockwise (CW) and all we have to do is tack on the direction at the end.
5. On a graph of speed vs time, constant positive acceleration will appear as

b. a straight line sloping upwards.

 
6. The scanner dish of an antenna completes 110 revolutions in one hour. Find the angular speed in:

A. RPM
B. rad/s

  
Knowns Unknowns Equations/Remarks
q = 110 rev
t = 1 hr
1 rev = 2p rad
1 hr = 60 min
1 min = 60 s		 A) w = ? in RPM

B) w = ? in rad/s

 w = q/t
A) w = q/t
    w = [(110 rev)/(1 hr)](1 hr/60 min)
    w = 1.83 RPM

B) w = (1.833 rev/min)(2p rad/1 rev)(1 min/60 s)
    w = 0.192 rad/s

A) Substitute in the known values, convert the hour to minutes, and calculate the answer.

B) Convert the RPM to rad/s.

We could go back to our original given values and find the answer in rad/s, but this involves a little more work, and the answer is the same when rounded to three significant figures.

7. An electric motor shaft starts from rest and reaches its design speed of 960 RPM 2.7 seconds after the motor is started. Find:

A. The angular acceleration in rad/s2
B. The average angular speed during acceleration
C. The total number of revolutions the motor turned through during acceleration

  
Knowns Unknowns Equations/Remarks
wi = 0 RPM
wf = 960 RPM
t = 2.7 s
1 rev = 2p rad
1 min = 60 s		 A) a = ? in rad/s2

B) wavg = ?

C) q = ? in rev

 a = (wf - wi)/t

wavg = (½)(wf + wi)

w = q/t

A) a = (wf + wi)/t
    a = [(960 rev/min - 0 rev/min)/(2.7 s)](2p rad/1 rev)(1 min/60 s)
    a =  37.1 rad/s2

B) wavg = (½)(wf + wi)
    wavg = (1/2)(960 rev/min + 0 rev/min)
    wavg = 480 rev/min

C) w = q/t
    q = wavgt
    q = (480 rev/min)(2.7 s)(1 min/60 s)
    q = 21.6 rev

A) Substitute the known values in the angular acceleration equation, convert revs to rads and minutes to seconds, and calculate the answer.

B) Substitute the known values in the average angular speed equation. The problem didn't ask for any particular units for part B, so leave the answer in rpm (or, if you want, convert the minutes to seconds now and get rev/s) since part C asks for revolutions.

C) Rearrange the angular speed equation to solve to the angle. Convert minutes to seconds.
8. Define the following rates with their unit: angular velocity and angular acceleration.

Angular velocity: the rate of change of angle with time (w = Dq/Dt), units in rad/s or rpm.

Angular acceleration: the rate of change of angular velocity with time (a = Dw/Dt), units in rad/s2.

 
9. A flywheel of radius 60 cm, initially at rest, requires 25 seconds to reach an angular velocity of 1750 rpm. Find the angular acceleration of the flywheel.  
Knowns Unknowns Equations/Remarks
r = 60 cm
wi = 0 RPM
wf = 1750 RPM
t = 25 s
1 rev = 2p rad
1 min = 60 s 		a = ? a = (wf - wi)/t
a = (wf - wi)/t

a = [(1750 rev/min - 0 rev/min)/(25 s)](2p rad/1 rev)(1 min/60 s)

a = 7.33 rad/s2

Substitute the known values in the angular acceleration equation. Convert revs to rads and minutes to seconds and calculate the answer.

Notice that the radius was extra information that was not needed so solve this problem.
10. A motor rotates at 5000 rpm. How much time is required for one revolution?  
Knowns Unknowns Equations/Remarks
w = 5000 RPM
q = 1 rev
1 min = 60 s 		t = ? w = q/t
w = q/t

t = q/w

t = [(1 rev)/(5000 rev/min)](60 s/1 min)

t = 0.012 s

Rearrange the angular speed equation to solve for the time. Convert minutes to seconds. Calculate the answer.
11. What is the initial angular velocity of a flywheel if, after two minutes, the flywheel is rotating clockwise at 250 rad/s and the acceleration is a constant 4 rad/s2 clockwise?  
Knowns Unknowns Equations/Remarks
wf = 250 rad/s CW
t = 2 min
a = 4 rad/s2 CW
1 min = 60 s 		wi = ? a = (wf - wi)/t
a = (wf - wi)/t

wf - wi = at

wi = wf - at

wi = (250 rad/s CW) - (4 rad/s2 CW)(2 min)(60 s/1 min)

wi = –230 rad/s CW, therefore:

wi = 230 rad/s CCW

Rearrange the angular acceleration equation to solve for the initial angular velocity.

Substitute in the known values, convert minutes to seconds, and calculate the answer.

The result is a negative value clockwise. The negative of a vector has the same magnitude but the opposite direction. The answer, then, is 230 rad/s CCW.
12. Through what angular distance does a flywheel turn while accelerating from 10 rad/s to 100 rad/s if it undergoes a constant acceleration of 14 rad/s2?

The acceleration is constant, so we can use the average angular speed to find the angle through which the flywheel turned while it was accelerating. We need to find the time it took to accelerate from the initial angular speed to the final angular speed. We can do this by rearranging the angular acceleration equation to find the time. Then we'll have enough information to find the angle.

 
Knowns Unknowns Equations/Remarks
wi = 10 rad/s
wf = 100 rad/s
a = 14 rad/s2 		q = ? w = q/t
a = (wf - wi)/t
wavg = (½)(wf + wi)
wavg = (½)(wf + wi)
wavg = (1/2)(100 rad/s + 10 rad/s)
wavg = 55 rad/s

a = (wf - wi)/t
t = (wf - wi)/a

wavg = q/t
q = wavg · t

q = (wavg)(wf - wi)/a
q = (55 rad/s)(100 rad/s - 10 rad/s)/(14 rad/s2)
q = 354 rad

Let's start off by finding the average angular speed. Substitute the known values in the equation and calculate.

Now we need to find the time it took to accelerate the flywheel. We can do this by rearranging the angular acceleration equation to find the time. I'm leaving it in symbolic form for right now for later substitution.

We now have the average angular speed and the time, so we can find the angle through which the flywheel rotated. Rearrange the angular speed equation to solve for the angle.
13. If the linear velocity of a point on a flywheel is v, then the angular velocity of that point is v/r, where r is the radius out to that point. A car accelerates at 100 mi/hr2 around a curve that has a radius of 50 ft. Just prior to accelerating the car speed is 30 mph. What is the angular speed of the car at it leaves the curve if it takes 10 seconds to complete the curve?

The angular speed of the car as it left the curve would be its final speed divided by the radius of the curve (notice that this is what we derived in Problem #1). We have the car's initial speed, its acceleration, and the time it took to go through the curve, so we can use the angular acceleration equation to find the car's final velocity. The difficult part of this problem is converting miles and hours to feet and seconds.

 
Knowns Unknowns Equations/Remarks
a = 100 mi/hr2
vi = 30 mi/hr
t = 10 s
r = 50 ft
1 hr = 3600 s
1 mi = 5280 ft		 w = ? w = v/r
a = (vf - vi)/t
a = (vf - vi)/t
vf = vi + at
vf = (30 mi/hr) + (100 mi/hr2)(10 s)(1 hr/3600 s)
vf = (30.28 mi/hr)(5280 ft/1 mi)(1 hr/3600 s)
vf = 44.41 ft/s

w = vf/r
w = (44.41 ft/s)/(50 ft)
w = 0.888 rad/s

Rearrange the linear acceleration equation to find the speed of the car as it exits the curve. Normally I would leave it in symbolic form and substitute that in the next equation, but the unit conversions in this problem are pretty lengthy. In fact, I broke the conversion in half so it would be easier to follow.

Substitute the final velocity and the radius in the last equation to find the angular speed of the car as it leaves the curve. Remember to put in the radians.