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.)

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 ?

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

To solve this problem, we can use the kinematic equations for angular motion. The relevant equations are:

1) Tangential acceleration: at = α * r
2) Tangential velocity: vt = ω * r
3) Angular acceleration: α = Δω / Δt
4) Radial (centripetal) acceleration: ar = (v^2) / r

Let's solve the problem step by step:

A) To find the magnitude of the rock's angular acceleration at t = 1.60 s, we need to find the change in angular velocity (Δω) and the change in time (Δt). The given information tells us the tangential deceleration (at = -1.40 m/s^2), the initial tangential velocity (vt = 3.00 m/s), and the radius of the wheel (r = 57.0 cm).

Using equation (1), we can find the angular acceleration:
at = α * r
-1.40 m/s^2 = α * 0.57 m

Simplifying the equation, we find:
α = -1.40 m/s^2 / 0.57 m
α ≈ -2.46 rad/s^2

Therefore, the magnitude of the rock's angular acceleration at t = 1.60 s is approximately 2.46 rad/s^2.

B) To find the time at which the magnitude of the rock's acceleration is equal to g (acceleration due to gravity), we need to consider both the tangential and radial accelerations. The radial acceleration (ar) is given by equation (4).

Since g is given as -9.8 m/s^2 (assuming downward direction is positive), we have:
ar + at = g
(v^2) / r + α * r = g

Substituting the given values into the equation:
(3.00 m/s)^2 / 0.57 m + α * 0.57 m = -9.8 m/s^2

Simplifying the equation, we find:
9.878 m^2/s^2 + α * 0.57 m = -9.8 m/s^2

Rearranging the equation, we have:
α = (-9.8 m/s^2 - 9.878 m^2/s^2) / 0.57 m
α ≈ -35.34 rad/s^2

Now, using equation (3), we can find the change in time:
α = Δω / Δt
Δω = α * Δt
Δt = Δω / α

Plugging in the values to get the change in time:
Δt ≈ 1.60 s / -35.34 rad/s^2
Δt ≈ -0.0452 s

Since time cannot be negative, we ignore this negative solution. Therefore, there is no real solution for which the magnitude of the rock's acceleration is equal to g.

In conclusion:
A) The magnitude of the rock's angular acceleration at t = 1.60 s is approximately 2.46 rad/s^2.
B) There is no time at which the magnitude of the rock's acceleration is equal to g.