part c: doesn't it rotate 2Pi in one period?
a: How did you get it 3 rev per 2 seconds? that is not given.
B, of course is wrong because of period is wrong.
A bike wheel rotates uniformly through 2 revolutions in 3 seconds.
a) what is the period of the wheel's rotation?
T=3rev/2s=1.5s
b) what is the frequency of the wheel's rotation? (in Hertz)
f=1/T
=1/1.5
=.666Hz
c) what is the angular speed of the wheel(provide answer in degrees, radians and revolutions)
How would I figure out part c? also did I do a) and b) correctly?
4 answers
Well isn't T = to the time of one revolution? So I took 3rev/2s to get 1.5s
Tammmy. IT did not complete 3 revolutions. Reread the problem.
Bob is right, it did not complete 3 revolutions, but the answer for a) is still correct, for the wrong reason.
Wheel completes 2 revolutions in 3 seconds. Period is amt of time to complete 1 revolution. Think of it as a distance problem.
Distance (# of revolutions) = Rate (revolutions/second) * Time (seconds)
So 1 revolution = (2 revolutions per 3 seconds) * Time
Time = (1 rev)/(2 revs/3 seconds) = 3/2 seconds, or 1.5 seconds
For part C, just divide the distance around the circle by the period 1.5s
i) 360 degrees/1.5 sec = 240 degrees/sec
ii) 2pi radians/1.5 sec = (4/3)pi rad/sec
iii) 1 revolution/1.5 sec = 2/3 rev/sec
Wheel completes 2 revolutions in 3 seconds. Period is amt of time to complete 1 revolution. Think of it as a distance problem.
Distance (# of revolutions) = Rate (revolutions/second) * Time (seconds)
So 1 revolution = (2 revolutions per 3 seconds) * Time
Time = (1 rev)/(2 revs/3 seconds) = 3/2 seconds, or 1.5 seconds
For part C, just divide the distance around the circle by the period 1.5s
i) 360 degrees/1.5 sec = 240 degrees/sec
ii) 2pi radians/1.5 sec = (4/3)pi rad/sec
iii) 1 revolution/1.5 sec = 2/3 rev/sec
3 answers