Calculating Revolutions with Constant Force: Solving for Angular Velocity

In summary, the disk has 1.0m of string wound around it. It is pulled off the disk by a constant force in 4.9s. The disk makes 4.822 revolutions while the string is being pulled off.
  • #1
blackbyron
48
0

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.
 
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  • #2
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 = Lf - Li = Lf = Iωf
f = (I/r)Δv
ωf = Δv/r, Δv = vf - vi = vf = Δs/Δt = 1 m/4.9 seconds

Then, θ = (ωf - ωi)*Δt = ωf*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.
 
  • #3
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.
 

1. How does disk pulling by the string work?

The principle behind disk pulling by the string is that the string exerts a force on the disk, causing it to move. This force is known as tension and is created by the string being pulled taut.

2. What materials are needed for disk pulling by the string?

You will need a disk with a hole in the center, a string or rope, and a source of tension, such as a weight or another person pulling on the other end of the string.

3. Can disk pulling by the string be used for transportation?

No, disk pulling by the string is not a practical method of transportation as it requires a constant source of tension and is usually only effective over short distances.

4. Is there a limit to how much weight can be pulled by a string?

Yes, there is a limit to how much weight can be pulled by a string. This limit is determined by the strength of the string and the amount of tension it can withstand before breaking.

5. Are there any safety precautions to take when performing disk pulling by the string?

Yes, it is important to ensure that the string is not frayed or damaged and that it is securely tied to the disk and the source of tension. It is also recommended to wear gloves to protect your hands from potential friction burns from the string.

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