Posted by **Steph** on Friday, November 26, 2010 at 5:02pm.

A rock stuck in the tread of a 57.0 cm diameter bicycle wheel has a tangential speed of 3.00 m/s. When the brakes are applied, the rock's tangential deceleration is 1.40 m/s^2.

A) What is the magnitudes of the rock's angular acceleration at t=1.60 s?

B) At what time is the magnitude of the rock's acceleration equal to g? (Remember that you need to include both tangential and radial accelerations in computing the magnitude of the rock's acceleration.)

- physics -
**Damon**, Friday, November 26, 2010 at 5:21pm
wr = v

a = r dw/dt

1.4 = r dw/dt

dw/dt = angular acceleration = 1.4/r = 2.8/.57

assume constant angular deacceleration

radial acceleration = w^2 r

tangential acceleration = r dw/dt

(we know dw/dt from above)

w r = v = 3 m/s at t = 0

wo = 3/r

w = wo - (dw/dt)t

that gives you w^2 r as a function of time

so when is

(w^2 r )^2 + (r dw/dt)^2 = 9.8^2 ?

- physics -
**Anonymous**, Monday, November 21, 2011 at 11:08pm
big dicks.

- physics -
**...**, Monday, November 28, 2011 at 4:45am
why does damons answer have to make no sense -_- im sure its right but its so confusing the way its written

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