# A dial can spin on a fixed rotational axis

• Danielheidarr
In summary: 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 use that representation to solve for the quantities.Yes, I understand that. But what if the number 24.52 appears from nowhere?If the number 24.52 appears from nowhere, we would need to find another way to solve for the quantities.
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

Danielheidarr said:
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.

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

Danielheidarr said:
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...

topsquark

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

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 and topsquark
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?

Danielheidarr said:
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?

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?

Danielheidarr said:
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.

berkeman
Danielheidarr said:
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.

berkeman
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.

PeroK said:
... 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.##

Steve4Physics and PeroK
PeroK said:
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.

Danielheidarr said:
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.

PeroK

## 1. How does a dial spin on a fixed rotational axis?

A dial can spin on a fixed rotational axis due to the presence of a pivot point or bearing at the center of the dial. This allows the dial to rotate freely around the axis without any lateral movement.

## 2. What is the purpose of a fixed rotational axis on a dial?

The fixed rotational axis on a dial allows for precise and consistent rotation, making it easier to measure and adjust values on the dial. It also prevents any lateral movement that could affect the accuracy of the measurements.

## 3. Can a dial spin on a fixed rotational axis in both directions?

Yes, a dial can spin on a fixed rotational axis in both clockwise and counterclockwise directions. The direction of rotation depends on the force applied to the dial and the design of the pivot point or bearing at the center of the dial.

## 4. How does the fixed rotational axis affect the accuracy of the dial?

The fixed rotational axis plays a crucial role in maintaining the accuracy of the dial. It prevents any lateral movement that could introduce errors in the measurements. Additionally, the pivot point or bearing at the center of the dial allows for smooth and consistent rotation, ensuring precise readings.

## 5. Are there any limitations to a dial spinning on a fixed rotational axis?

While a dial can spin on a fixed rotational axis with great precision, there are limitations to the amount of weight or force it can handle. Exceeding these limitations can cause the dial to become unbalanced and affect its accuracy. Additionally, external factors such as friction and wear on the pivot point or bearing can also impact the dial's performance over time.

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