Posted by **andrew** on Monday, October 4, 2010 at 5:29pm.

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.

- Physics -
**bobpursley**, Monday, October 4, 2010 at 5:33pm
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.

- Physics -
**andrew**, Monday, October 4, 2010 at 5:35pm
Ok thanks.

- Physics -
**Mak**, Saturday, October 27, 2012 at 1:49pm
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

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