# revolutions with constant angular decelleration

[Linus Pauling]

1. A well-lubricated bicycle wheel spins a long time before stopping. Suppose a wheel initially rotating at 120 rpm takes 65 s to stop.

If the angular acceleration is constant, how many revolutions does the wheel make while stopping?



2. a = omega^2*r



3. I already know the solution is theta = 65 revolutions. What is the calculation?

[hiuting]

ok so:

120rpm = 12.57rad/s << obtained by dimensional analysis 'cause you know 1 rev = 2(pie)rads

12.57 rad/s is your initial omega

your final omega = 0 rad/s

using the kinematics equation : omega final = omega initial + alpha (delta time)
you get : 0 = 12.57 + (alpha)(65s)

-0.1934rad/s^2 = alpha << angular deceleration

so, using another kinematics equation, the one with the angular displacement:
final position = initial position + 12.57(65) + (0.5)(-0.1934)(65^2)
then you get 408.4925 rads
as a result, (use the dimensional analysis again) to get the revolutions! which is 65.01 revolutions.

i hope that helped. i might be a tad confusing :S

[Linus Pauling]

I understand the first part, and the reasoning used in the second, but I don't understand this part:

-0.1934rad/s^2 = alpha << angular deceleration

[hiuting]

I understand the first part, and the reasoning used in the second, but I don't understand this part:

-0.1934rad/s^2 = alpha << angular deceleration

you know the kinematics formula : omega final = omega initial + (alpha)(time) ?

it's the same as
final angular velocity = initial angular velocity + (angular acceleration)(time)
since you have the final angular velocity and the initial, and the time, you can figure out the angular acceleration. :)

and ... why use : a = omega^2*r?
that formula is for the centripedal acceleration, and NOT the angular acceleration. centripedal is a linear acceleration! :)