Help with rotation of rigid bodies

AI Thread Summary
The discussion centers on calculating the velocity of a cylinder attached to two metal disks with different radii and masses, mounted on a frictionless axis. The total moment of inertia for the disks is correctly identified as 2.25 x 10^-3 kgm^2. When attempting to find the velocity of the cylinder just before it hits the ground, the user struggles with the kinetic energy calculations, particularly in accounting for the mass of the disks and the correct moment of inertia. It is emphasized that the final kinetic energy equation must include the moment of inertia of both disks and the cylinder, and all units should be consistent. Properly applying these principles will yield the correct velocity results.
RussG
Messages
3
Reaction score
0
Alright, I'm stuck on this problem.

Two metal disks, Radius 1 = 2.5cm, Mass 1 = 0.8kg and Radius 2 = 5.0cm, Mass 2 = 1.6kg are wielded together and mounted on a frictionless axis through their common center.

The first part asked what is the total moment of inertia of the two discs, which I got and it was 2.25 x 10^-3 kgm^2

Part b attaches a string to the smaller disk holding a cylinder of mass 1.5kg 2m above the ground. It asks the velocity of the cylinder right before it hits the ground when it's released from rest.

Then it moves the string in part c to the larger disc. I have the solutions but am not getting the same answers when I attempt to solve for it. I know that when attached to the larger disc it will have a greater speed because of the greater mass since their radiuses do not matter.

I have tried using the good ol K1 + U1 = K2 + U2 to determine its velocity.

I know K1 at rest is 0, and U1 is mgh. I know U2 is 0. So I need the final kinetic energy, right? Well the final kinetic energy would be 1/2mv^2 + 1/2Iw^2, or am I getting something wrong here?

since w = v/R I substituted that in.

I get 1/2(1/2MR^2)(v/R)^2 + 1/2mv^2 = mgh

solving for v I get sqrt((2gh)/(1+M/2m)) = v

Plugging in the numbers I'm not coming up with the correct velocities. Where have I gone wrong? Thanks for any help!
 
Physics news on Phys.org
You have almost got it correct! You need to take into account the mass of the disks when calculating the final kinetic energy. The equation for final kinetic energy should be: K2 = 1/2(M1 + M2 + MR^2)(v/R)^2 + 1/2mv^2 Where M1 and M2 are the masses of the two disks and MR^2 is the moment of inertia of the two disks. Once you have this equation, you can proceed as you did before to find the velocity of the cylinder right before it hits the ground.
 




Hi there,

It seems like you are on the right track, but there may be a few errors in your calculations. First, when using the equation K1 + U1 = K2 + U2, make sure that you are consistent with your units. In this case, since the masses were given in kilograms, make sure to use joules for energy and meters for height.

Next, when calculating the final kinetic energy, you should use the moment of inertia of the entire system (both disks and the cylinder attached) since they will all be rotating together. So the equation would be 1/2(I1 + I2 + Icyl)v^2, where Icyl is the moment of inertia of the cylinder about the axis of rotation.

Also, when substituting w = v/R, make sure to use the radius of the entire system, which would be the sum of the radii of the two disks.

Lastly, when plugging in numbers, make sure to convert all units to their base units (e.g. cm to m) to get consistent results.

I hope this helps and good luck with your problem!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top