Calculating Revolutions with Constant Force: Solving for Angular Velocity

[blackbyron]

Homework Statement

A solid disk with a radius of 3.3cm is a rest. The disk has 1.0m of string wound on to its circumference. The string is pulled off the disk by a constant force in a time of 4.9s. How many revolutions dose the disk make while the string is being pulled off?

Homework Equations

1 rev = 2*pi

V =(2*pi*r)/t

The Attempt at a Solution

find velocity.
Do I need to start off using energy converstion?

(1/2)(m)(v^2) = m*g*x

and I'm lost here

Any help will be appreciated.

 

[da_nang]

Assuming that the disk is spinning on the z-axis with no friction, consider that M = dL/dt = I*dω/dt, where M is torque, L is the angular momentum and I is the moment of inertia.

The string is pulled with a constant force F on a distance s = 1m in 4.9 seconds.

M = I*dω/dt = I*d(v/r)/dt = (I/r)dv/dt
Since dv/dt = a, and it's a constant acceleration due to a constant force, then:
M = (I/r)Δv/Δt
MΔt = (I/r)Δv
Remember, M = dL/dt = ΔL/Δt, since the change is constant.

(ΔL/Δt)Δt = (I/r)Δv
ΔL = (I/r)Δv, ΔL = L_f - L_i = L_f = Iω_f
Iω_f = (I/r)Δv
ω_f = Δv/r, Δv = v_f - v_i = v_f = Δs/Δt = 1 m/4.9 seconds

Then, θ = (ω_f - ω_i)*Δt = ω_f*Δt = (Δv/r)Δt = (ΔsΔt)/(rΔt) = Δs/r

Revolutions = θ/(2pi) = Δs/(2r*π) = 1 m /(2*0.033 m * π) = 4.822... revs.

Another way to see it is that, since the string is wound around the disk, the disk's circumference must also spin the same distance the string would need to travel. In other words, if the string travels 1 m, the circumference travels 1 m.

 

[blackbyron]

I see you're using inertia to solve, don't know why my teacher didn't talk about that on angular acceleration.

But Thanks, tho, I'll take some time to look at it.