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.

Question

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.)

Answer 1

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 ?

Answer 2

why does damons answer have to make no sense -_- im sure its right but its so confusing the way its written