A disk with a Spring around it

[lebprince]

Homework Statement

A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.05 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.09 m, each with a small mass m = 0.7 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 27 N. At the instant when the center of the disk has moved a distance d = 0.038 m, a length w = 0.020 m of string has unwound off the disk.


(a) At this instant, what is the speed of the center of the apparatus?
v = m/s

(b) At this instant, what is the angular speed of the apparatus?

(c) You keep pulling with constant force 27 N for an additional 0.035 s. Now what is the angular speed of the apparatus?



I have no idea how to approach the problem. any help please

 

[Hootenanny]

Well, a good place to start would be to determine the moment of inertia of the object.

 

[lebprince]

ok so the moment of inertia of the object is 1/2Mr^2 so = 1/2*2.5*(0.05)^2 = 0.003125

 

[Hootenanny]

ok so the moment of inertia of the object is 1/2Mr^2 so = 1/2*2.5*(0.05)^2 = 0.003125

That we be the moment of inertia of the disk. What about the the small masses?

 

[lebprince]

That we be the moment of inertia of the disk. What about the the small masses?

ok since we have 4 masses ; m1r^2 + m2r^2 + m3r^2 + m4r^2

so 0.2268

 

[Hootenanny]

ok since we have 4 masses ; m1r^2 + m2r^2 + m3r^2 + m4r^2

so 0.2268

Correct. So the total moment of inertia is the sum of this and the moment of inertia of the disc.

What do you suppose the next step will be?

 

[lebprince]

Correct. So the total moment of inertia is the sum of this and the moment of inertia of the disc.

What do you suppose the next step will be?

umm am assuming i have to find the angular speed cause it would one one way to find V, but i know the Lrot = IW but i don't have Lrot

 

[Hootenanny]

Consider the force and resultant accelerations of the body.

 

[lebprince]

Consider the force and resultant accelerations of the body.

so i can use F = mxa?

 

[Hootenanny]

so i can use F = mxa?

Yes, but you also need to consider the torque on the body.

 

[lebprince]

Yes, but you also need to consider the torque on the body.

ok..so i can find the torque on the body using r x Fnet, now is r the distance the center of the disk has moved? so 0.038 x 27 = 1.026? and F = mxa so a = f/m = 27/2.5 = 10.8

 

[Hootenanny]

ok..so i can find the torque on the body using r x Fnet, now is r the distance the center of the disk has moved? so 0.038 x 27 = 1.026?

No, r is the [perpendicular] distance from the point of action of the force causing the torque and the centre of rotation.

 

[lebprince]

No, r is the [perpendicular] distance from the point of action of the force causing the torque and the centre of rotation.

ok so i would use 0.05 x 27 = 1.35 and F = mxa so a = f/m = 27/2.5 = 10.8