A dial can spin on a fixed rotational axis

[Danielheidarr]

Homework Statement

A dial can spin on a fixed rotational axis. The mass of the dial is m. On the right side of the dial is a little box also with the mass m, but on the left side of the dial there is a little box with a mass of 1/2 m. The gravity of the earth pulls of course harder on the right side of the dial, so when the dial is released it starts turning clockwise. The moment of inertia is Id=1/2mR², where R is the radius of the dial.

Relevant Equations

a) what is the angular velocity of the dial just after it has been released?
b) Use conservation of energy to find the angular velocity of the system when the heavier mass is at the lowest point and the lighter on at the highest point.
c) What is the angular velocity of the system in the same state as in b)?

What I have done is on my Ipad that I cant upload or at least don't know how to... :/
With hope of help
DJ

 

[berkeman]

What I have done is on my Ipad that I cant upload or at least don't know how to... :/

Welcome to PF. Try the "Attach files" link below the Edit window. It's best to type your work into the PF post using LaTeX (see the "LaTeX Guide" link below the Edit window, but for now if your work is legible, start by attaching it so we can look it over. Thanks.

 

[Danielheidarr]

Yeah I'm not using LaTeX right now I'm just drawing and writing on my Ipad.

 

[berkeman]

Yeah I'm not using LaTeX right now I'm just drawing and writing on my Ipad.

We do require math to be posted in LaTeX (and it's easy to learn the basics via that link below), but for now since you're new, let's have a look at what you've done on your iPad...

 

[Danielheidarr]

 

[Danielheidarr]

I finally found out how to do it, but the language is probably something you don’t understand but so be it

 

[berkeman]

Yeah, my translation skills (especially from cursive) are not very good today...

But it does look like you are saying for part a) that $\omega_0$ is not zero? I would expect the initial angular velocity to be zero as the wheel is released, no? And $\omega$ will not be zero when the heavier weight is at the bottom of the spin for part c), will it?

 

[Danielheidarr]

in a) I have to find omega just after it is released would it still be zero then?
and in c) if it would be released in that state why would it not be zero?

 

[berkeman]

in a) I have to find omega just after it is released would it still be zero then?
and in c) if it would be released in that state why would it not be zero?

The way I interpret the problem, the motion of the disc will be an oscillation (assuming no axle friction), where the larger mass starts at the 3 o'clock position, swings down and through 6 o'clock, and up to the 9 o'clock position where it stops and goes back down again. Do you interpret this motion in a different way?

 

[Danielheidarr]

Well.. I guess that would make sense. I just feel like the question needs more information. But lets say we use you idea why would omega in a) be 0 if we are going to check it just after it is released?

 

[PeroK]

Well.. I guess that would make sense. I just feel like the question needs more information. But lets say we use you idea why would omega in a) be 0 if we are going to check it just after it is released?

Initial angular velocity is $0$ as the disc is initially at rest. Initial angular acceleration is not zero.

 

[kuruman]

Well.. I guess that would make sense. I just feel like the question needs more information. But lets say we use you idea why would omega in a) be 0 if we are going to check it just after it is released?

Why not? If you hold a rock above ground, what is the rock's velocity as soon as your fingers let go? You are asked this question to make sure that when you answer part (b), you use the correct initial kinetic energy.

 

[haruspex]

There are two reasons the rules say that images are for diagrams and textbook extracts, not working.
1. The images are frequently hard to read - faint, blurred, poor handwriting, upside down…
2. It makes it hard to refer to specific parts of the working in making comments.

But I can read that you started with the equation v=mg=(m+m/2)g.
1. mg would be a force, and a force cannot equal a velocity. A very good habit to get into is to check equations for consistent dimensions.
2. never use the same symbol for two different things in an equation (“m=m+m/2”), and preferably not in the same solution. Use subscripts to distinguish.
3. On what basis would you add those two masses anyway?

 

[PeroK]

... there is also the issue that the number 24.52 appears from nowhere.

 

[kuruman]

... there is also the issue that the number 24.52 appears from nowhere.

Ignoring units and for $g=9.81$, it should be obvious that $\frac{(m+m+\frac{1}{2}m)g}{m}=24.52.$

 

[Danielheidarr]

Initial angular velocity is $0$ as the disc is initially at rest. Initial angular acceleration is not zero.

what would the initial angular acceleration be. how do we find that out when we don't have any numbers given.

 

[kuruman]

what would the initial angular acceleration be. how do we find that out when we don't have any numbers given.

We find a symbolic representation, $\alpha = \dots$, in terms of the quantities given, namely $m$, $R$ and the usual constants like $g$, $\pi$, etc. If numbers for these are given, we can find a number for the initial acceleration; if no numbers are given, we leave the expression as is. Note that you don't need the angular acceleration to answer this question because the problem is asking you to use mechanical energy conservation.