Moment of Inertia problem with multiple masses

[amcelroy13]

Homework Statement

A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.04 m. Attached to the disk are four low-mass rods of radius b = 0.09 m, each with a small mass m = 0.6 kg at the end in an "x" configuration through the center of the disc. What is the moment of inertia for the object?

Homework Equations

I for small masses = r1^2(m1) + (r2^2)m2)...etc
I for disk = (1/2)MR^2

The Attempt at a Solution

This is only part of a problem I'm doing. I feel as though the total I should be equal to the I of the disk + the I of the masses, which would mean the equation is "I = .5MR^2 + 4(mb^2)" but this is not giving me the correct answer. Any ideas?

 

[tiny-tim]

Welcome to PF!

Hi amcelroy13 Welcome to PF!

(try using the X² tag just above the Reply box )

I feel as though the total I should be equal to the I of the disk + the I of the masses, which would mean the equation is "I = .5MR^2 + 4(mb^2)" but this is not giving me the correct answer. Any ideas?

Your formula looks ok to me …

(but are you sure it says "radius" … rods don't usually have a radius?)

maybe you have the decimal point in the wrong place …

what figures are you getting?

 

[amcelroy13]

Ya, the shape is kinda strange, turns out my calculator was almost out of batteries so it kept giving me wrong answers. I had no idea that could happen. however, the next part is proving extremely difficult for me.

This is the actual, complete question:

A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.04 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 = 0.6 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 = 26 N. At the instant when the center of the disk has moved a distance d = 0.030 m, a length w = 0.011 m of string has unwound off the disk.

(a) At this instant, what is the speed of the center of the apparatus?
v = .564 m/s

(b) At this instant, what is the angular speed of the apparatus?
omega1 = 5.165 radians/s

(c) You keep pulling with constant force 26 N for an additional 0.039 s. Now what is the angular speed of the apparatus?
omega2 = ? radians/s

Necessary equations:
Lf = Li + T(deltat)
T = r x p (cross product)
Li = r x F (cross)

Attempt:
My biggest problem on this part is finding the initial angular momentum and how it relates to omega (angular speed). I feel as though it should just be "Li = R(total mass)(omega1)" which would make solving for omega two rather simple:
Lf = Li + T(dealtat)
R(total mass)(omega2) = R(total mass)(omega1) + R(F)(deltat)
omega2 = ((R(total mass)(omega1) + R(F)(deltat))/(R(total mass))
omega2 = (.04(4.9)(5.165) + .04(26)(.039))/(.04(4.9))
omega2 = 5.37

This however is not the correct answer, nor does it really make sense that it would speed up that fast in that short time interval. Any ideas?

 

[tiny-tim]

Ya, the shape is kinda strange, turns out my calculator was almost out of batteries so it kept giving me wrong answers.

My dog ate mine!

A string is wrapped around a uniform disk of mass M = 2.5 kg and radius R = 0.04 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 = 0.6 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 = 26 N. At the instant when the center of the disk has moved a distance d = 0.030 m, a length w = 0.011 m of string has unwound off the disk.
…

My biggest problem on this part is finding the initial angular momentum and how it relates to omega (angular speed). I feel as though it should just be "Li = R(total mass)(omega1)" which would make solving for omega two rather simple:
Lf = Li + T(dealtat)
R(total mass)(omega2) = R(total mass)(omega1) + R(F)(deltat)
omega2 = ((R(total mass)(omega1) + R(F)(deltat))/(R(total mass))
omega2 = (.04(4.9)(5.165) + .04(26)(.039))/(.04(4.9))
omega2 = 5.37

This however is not the correct answer, nor does it really make sense that it would speed up that fast in that short time interval. Any ideas?

oh, the centre of the disc isn't fixed!

(I'm finding it very difficult to read your work, so I'm not sure what you've done … have an omega: ω and a delta: ∆ )

ok, in (a) and (b) you found v and ω₁ from geometry … from that you should have been able to calculate the moment of inertia and then use that for (c).