What Temperature Allows a Ring to Slide Off a Sphere?

In summary, the problem involves a sphere with a ring around it and at a certain temperature the diameter of the sphere is 0.05% larger than the diameter of the ring. The question is at what temperature the ring will be able to come off the sphere. Attempting to relate the equations for the sphere and ring, the user gets stuck with too many unknowns. The equations for both objects involve the change in length being equal to a constant multiplied by the initial length and change in temperature. The user suggests using the alpha value for both objects as they are proportional to the diameter. However, the change in diameter for the sphere will be different than the change in diameter for the ring, so at some colder temperature when the diameters
  • #1
bpw91284
67
0
Problem
Sphere has ring around it. At T=70C the diameter of the sphere is 0.05% larger than the diameter of the ring. At what temp will the ring be able to come off sphere?
Attempt
D_s=1.0005*D_r at 70C

Sphere
delta_V=beta*V_o*delta_T
Ring
delta_L=alpha*L_o*delta_T

L=pi*D_r (circumference)
delta_L=alpha*(pi*D_r,o)*delta_T

Tried relating D_r and D_s with sphere volume eqn (V=4/3*pi*(D_s/2)^3) and circumference of ring eqn C=L_r=pi*D_r and I'm stuck...

Temeperature where D_r=D_s is the point when ring can come off, but I don't know how to get there.
 
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  • #2
use alpha for both because the size of sphere is proportional to the diameter. as the diameter gets larger or smaller the smaller the sphere's surface area gets. right? so you can use alpha for the shere and the right and make the lengths equal each other like you said.
 
Last edited:
  • #3
I just end up with too many unknowns.

Both
delta_L=alpha*L_o*delta_T
L=pi*D
Sphere
Eqn 1
pi*delta_D_s=alpha*(D_o,s)*delta_T
Ring
Eqn 2
pi*delta_D_r=alpha*pi*(D_o,r)*delta_T
Eqn3
D_o,s=1.0005*D_o,r
Eqn 4
D_f,r=D_f,s

Four eqns, 5 unknowns which are...
D_o,s
D_f,s
D_o,r
D_f,r
T_f
 
  • #4
what material is the sphere and what material is the ring? What does that tell you about alpha value for each one? also on started out at what length in regards to the other?
The change plus the initial value for each material should equal each other right?
 
  • #5
alphas don't matter, they are just from table. The only given info is that the diameter of the sphere is 0.05% larger than the diameter of the ring at the initial temp of 70C.

"The change plus the initial value for each material should equal each other right?"

The change in diameter of the sphere will be different than the change in diameter of the ring. If not, the ring would never come off. The sphere will shrink at a faster rate than the ring and so at some colder temp, when D_r=D_s, the ring can come off. I know the answer is 41C if that helps you work backwards.
 
  • #6
bpw91284 said:
alphas don't matter, they are just from table. The only given info is that the diameter of the sphere is 0.05% larger than the diameter of the ring at the initial temp of 70C.

"The change plus the initial value for each material should equal each other right?"

The change in diameter of the sphere will be different than the change in diameter of the ring. If not, the ring would never come off. The sphere will shrink at a faster rate than the ring and so at some colder temp, when D_r=D_s, the ring can come off. I know the answer is 41C if that helps you work backwards.

alpha's do matter, even if they are from a table. If they both had same materials, hence same alphas, that ring will never come off! Length initial of sphere equals .0005 times Length initial of ring added to length initial of ring. can you take it from here?
 
  • #7
Antineutron said:
alpha's do matter, even if they are from a table. If they both had same materials, hence same alphas, that ring will never come off! Length initial of sphere equals .0005 times Length initial of ring added to length initial of ring. can you take it from here?

You are repeating what I have already said…

The alpha comment, duh…
Your initial length sentence… my first post I stated that D_s=1.0005*D_r…

And no, I can't take it from here because you have not told me anything I didn't already know.
 

1. What is thermal expansion?

Thermal expansion is the tendency of matter to change in volume in response to a change in temperature.

2. Why does thermal expansion occur?

Thermal expansion occurs because as the temperature of a substance increases, the particles that make up the substance gain energy and vibrate more, causing them to take up more space.

3. How does thermal expansion affect different materials?

Different materials have different coefficients of thermal expansion, which means they expand at different rates when exposed to the same change in temperature. For example, metals tend to expand more than non-metals.

4. What are some practical applications of thermal expansion?

Thermal expansion has many practical applications, such as in the construction of bridges and buildings, where materials must be able to expand and contract without causing structural damage. It is also used in the creation of thermometers and thermostats.

5. How is thermal expansion measured?

Thermal expansion is typically measured using a device called a dilatometer, which measures the change in dimensions of a material as it is heated or cooled. The coefficient of thermal expansion can then be calculated using this data.

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