# Using the Conservation of Energy to find the Radius of a Bowl

• Alexandra Fabiello
In summary: If the table is perfectly level, the ball does not roll. If you tilt the table, it rolls.The ball will gain speed as it rolls to the edge of the table, depending on how far the table is tilted.
Alexandra Fabiello
Homework Statement
A small ice cube is released from rest at the top edge of a hemispherical bowl. When it reaches the bottom of the bowl, its speed is 2.5 m/s. Use conservation of energy to find the radius of the bowl in centimetres. Neglect friction.
Relevant Equations
1/2 mv1^2 + mgy1 = 1/2 mv2^2 + mgy2
v1 = 0 m/s
v2 = 2.5 m/s
y1 - y2 = distance a quarter of the way around the bowl (since we're neglecting friction)

mass can be factored out, so it isn't needed, and some simplifying and the like gets this formula:
v22 = v12 + 2g(y1 - y2)

so 2.52 = 0 + 2(9.8)(y)
6.25 = 19.6y
y = 0.318877551 m * 4 = 1.275510204 m = circumference of bowl = 2πr
1.275510204/2π = 0.203003753 m * 100 = 20 cm

But apparently this is wrong, so I don't know what got messed up.

Alexandra Fabiello said:
Homework Equations: 1/2 mv1^2 + mgy1 = 1/2 mv2^2 + mgy2
How are y1 and y2 defined in that equation?

haruspex said:
How are y1 and y2 defined in that equation?

I was using the start point at 0 m/s as y1, which is 0 m, and the end point at the bottom of the bowl as y2 at 2.5 m/s.

Alexandra Fabiello said:
I was using the start point at 0 m/s as y1, which is 0 m, and the end point at the bottom of the bowl as y2 at 2.5 m/s.
Your solution looked okay until you got hold of your calculator and went a bit crazy with decimal places.

What defines gravitational potential energy? And how does this relate to the radius of the bowl in this problem?

Did you draw a diagram?

Alexandra Fabiello said:
I was using the start point at 0 m/s as y1, which is 0 m, and the end point at the bottom of the bowl as y2 at 2.5 m/s.
That is not what I asked.
You quote a Homework Equation. It includes a number of variables, including y1 and y2.
An equation is meaningless without the definitions of the variables in it. In that equation, e.g., I understand that v1 and v2 are initial and final velocities of the mass m. How are y1 and y2 defined? Initial and final what?

Could have sworn I replied to this. Ah well.

I asked my teacher about this and she clarified some stuff. It's weird to think of displacement being the y1 and y2 and not distance traveled, especially since this means that the velocity direction should be given as the direction of displacement and not the direction of movement, which I got confused by.

I sort of get it now.

Alexandra Fabiello said:
displacement being the y1 and y2 and not distance traveled
In mechanics, displacement is a vector but distance is a scalar. Hence one can refer to e.g. vertical displacement, i.e. the vertical component.
Distance traveled is path length, whereas displacement is the final position vector minus the initial position vector.

To be clear, gravity doesn’t point sideways. There is no work done by gravity when moving an object across the force. The y in your gravitational potential energy mgy is always only the height dimension. How far the object moves in the other two dimensions doesn’t affect gravitational potential energy.

Alexandra Fabiello said:
Could have sworn I replied to this. Ah well.

I asked my teacher about this and she clarified some stuff. It's weird to think of displacement being the y1 and y2 and not distance traveled, especially since this means that the velocity direction should be given as the direction of displacement and not the direction of movement, which I got confused by.

So, what you would expect if you are standing on a sidewalk that is only just downhill at a small angle, then you'd go flying down that sidewalk just as though you'd jumped out of an aeroplane?

PeroK said:
So, what you would expect if you are standing on a sidewalk that is only just downhill at a small angle, then you'd go flying down that sidewalk just as though you'd jumped out of an aeroplane?

I mean, without friction maybe, I can't really visualize this stuff...

Alexandra Fabiello said:
I mean, without friction maybe, I can't really visualize this stuff...
That is a problem.

Alexandra Fabiello said:
I mean, without friction maybe, I can't really visualize this stuff...
Suppose that you have a ball and a table. Maybe it is a tennis ball on the kitchen table. Maybe it is a pool ball on a pool table.

If the table is perfectly level, the ball does not roll. If you tilt the table, it rolls.

The point being: If you tilt it further, it accelerates faster.

The speed it gains rolling to the edge does not depend only on how far it is to the edge of the table. It also depends on how far the table is tilted. The speed with which the ball rolls off the table will turn out to depend on the difference in height between where the ball started and where you measure its speed.

One can justify this with equations and vector addition and trigonometry. But it is handy to have a feel as well.

PeroK

## 1. How is the conservation of energy used to find the radius of a bowl?

The conservation of energy principle states that energy cannot be created or destroyed, but can only be transformed from one form to another. In the case of a bowl, the potential energy due to its height is converted into kinetic energy as an object rolls down the bowl. By equating the initial potential energy to the final kinetic energy, we can find the radius of the bowl.

## 2. What are the key equations used in this method?

The two main equations used are the conservation of energy equation: PE = KE, where PE is the potential energy and KE is the kinetic energy, and the equation for potential energy of an object at a given height: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

## 3. Is the mass of the object considered in this calculation?

Yes, the mass of the object is an important factor in determining the radius of the bowl. This is because the potential energy of an object is directly proportional to its mass, so a heavier object will have more potential energy and will require a larger radius to balance out the kinetic energy.

## 4. Can this method be used for any type of bowl?

Yes, this method can be used for any type of bowl as long as it has a smooth surface and the object is able to roll without slipping. This means that the frictional forces are negligible and the energy is conserved.

## 5. Are there any limitations to using the conservation of energy to find the radius of a bowl?

One limitation is that this method assumes a perfect bowl shape, which may not be the case in real life. Additionally, it does not take into account any external forces acting on the object, such as air resistance or imperfections in the bowl's surface. These factors can affect the accuracy of the calculation.

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