A bicycle wheel that initially spins at 3 revolutions per second is braked uniformly to a stop in 1 second. A spot is painted at the outer edge of the wheel. How many revolutions does the spot go during the time it takes the wheel to stop?
The answer I got was 0.5 but was incorrect so can someone show me how to do this problem.
I like moving things to radians, not revs.
angularaccleration=changevleociyt/time
= -3*2PI/1 rad/sec
finalangulvelocity^2=intialangularvelocity^2+2*angularacceleration*displacement
0=(3*2PI)^2+2(-2*3PI)*displacment
solve for displacement (in radians) ; divide by 2PI to get revs.
Ok thanks.
1.5 revolutions.
If it starts at 3 rev/sec and ends at 0 rev/sec and decelerates uniformly, then the avg is 1.5 rev/sec. It spent 1 second, so 1.5 rev/sec * 1 sec = 1.5 rev
To solve this problem, we need to understand the relationship between angular velocity (ω), time (t), and the number of revolutions (N).
The formula to calculate the number of revolutions is given by:
N = ω * t / (2π)
Where:
N = number of revolutions
ω = angular velocity (in radians per second)
t = time (in seconds)
In this case, the initial angular velocity (ω) is given as 3 revolutions per second. To convert this into radians per second, we need to multiply by 2π since there are 2π radians in one revolution. Therefore, the initial angular velocity is:
ω = 3 * 2π = 6π radians per second
Next, we know that the wheel comes to a stop in 1 second (t = 1).
Now, plugging the values into the formula, we can calculate the number of revolutions:
N = (6π) * 1 / (2π) = 3 revolutions
So, during the time it takes the wheel to stop, the spot painted at the outer edge of the wheel goes through 3 complete revolutions, not 0.5 as you initially thought.
Therefore, the correct answer is 3 revolutions.