Asked by andrew
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.
The answer I got was 0.5 but was incorrect so can someone show me how to do this problem.
Answers
Answered by
bobpursley
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.
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.
Answered by
andrew
Ok thanks.
Answered by
Mak
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
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