A small lump of ice sliding down a large, smooth sphere.

In summary, the ice block slides down the sphere with a velocity that is proportional to the cosine of the angle it has slid down. The normal force is proportional to the square of this velocity.
  • #1
MrDaahl
9
0

Homework Statement


A small lump of ice is sliding down a large, smooth sphere with a radius R. The lump is initially at rest. To get it started, it starts from a position slightly right to the sphere's top, but you can count it to start from the top. The lump is fallowing the sphere for a certain amount of time as shown in the figure. The sphere does not move.

http://i.tinyuploads.com/r6tkWK.png

2. Questions
a) Write an expression for the ice lump velocity when it is has slided down the angle.

b) Draw a force diagram on the block when it has slided down to the angle.

c) Determine the normal force size when the ice lump has completed the angle.

When the angle is large enough, the ice block no longer follow the sphere's surface, but fall freely.

d) Decide/Calculate this angle.

The Attempt at a Solution


So I am pretty sure i got the solution for queation a) :
When you guys said "Use conservation energy" i thought about using the radius to calculate the height of the circle. I puzzled abit with some formulas and got this equation:

mgh = 3/4mv^2

In question b) i made a FBD diagram to show where the forces are pointing. It looks like this:
http://i.tinyuploads.com/HSSjC4.png

The c) solution must be like this:
Fn = -FT
 
Last edited:
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  • #2
I wonder if conservation of energy wouldn't help...
 
  • #3
For part a) use conservation of energy

For part b) , make an FBD of the block .Mark all the forces acting on the block.Resolve the forces tangentially and radially.
 
  • #4
Ok, so I've to do something with R*2, which will equal the height of the circle

So i got this equation so far:

V1 = √(2*g*h+v02)

Can anyone confirm the equation?
 
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  • #5
Do you know how to apply conservation of energy ?
 
  • #6
Apparently not.. I assume

I thought it should be the block's potentiel + kinetic energy at the initialising state, equal the kinetic energy at the secondary state.
 
  • #7
What's the initial kinetic energy though?
 
  • #8
Isnt it zero?
 
  • #9
MrDaahl said:
Apparently not.. I assume

I thought it should be the block's potentiel + kinetic energy at the initialising state, equal the kinetic energy at the secondary state.

KE1+PE1=KE2+PE2

What is PE1 ? Assume h=0(reference of potential) to be at the center of the sphere .In other words measure heights from the center of sphere ,not the ground.
 
  • #10
What's the block's initial potential energy?
 
  • #11
I think OP was assuming that h=0 was at final position (so that potential energy could be zero).
 
  • #12
Tanya Sharma said:
KE1+PE1=KE2+PE2

What is PE1 ? Assume h=0(reference of potential) to be at the center of the sphere .In other words measure heights from the center of sphere ,not the ground.

That was wrong.

PE1 = mgR
 
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  • #13
MrDaahl said:
PE1 = 0.

Do you think the block is at zero height with respect to the center of the sphere ?
 
  • #14
If h=0 is the center of the circle, the block is never there, so h is never 0. so PE can't be 0, either. You can choose h=0 to be where PE1 is measured or where PE2 measured if you like, just make sure you do the rest of the problem keeping that coordinate system.
 
  • #15
KE1+PE1=KE2+PE2

KE1 = 0

PE1 = mgR

KE2 = ½mv2

PE2 = mgΔR.

→ mgR=½mv2+mgΔR

→ this doesn't make sense :cry:
 
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  • #16
if h is initially R, it is not finally 0... The vertical distance traveled by the block (h2-h1) is not equal to the radius of the circle. And you already told us that KE1 = 0 (as it should).
 
  • #17
I know the R is not the same value in the initialising state as it is in the next.
But i simly cannot see how to calculate PE_2
 
  • #18
You will have to use some trigonometry to see the vertical distance traveled. Your change in potential energy is going to be [itex]mg \Delta h[/itex], essentially.
 
  • #19
MrDaahl said:
I know the R is not the same value in the initialising state as it is in the next.
But i simly cannot see how to calculate PE_2
R is the radius of the sphere, that does not change.
Call the centre of the sphere O, the initial position of the ice A, and the later position B.
Draw a line horizontally from B to meet the line AO at C. You know the angle COB and the length OB. So what is length OC? Length AC?
 
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  • #20
wait
 
  • #21
The length OC must be 3/4 R and the length of AC must therefore be 1/4 R
 
  • #22
MrDaahl said:
The length OC must be 3/4 R and the length of AC must therefore be 1/4 R
No, it must depend on theta. How did you get 3R/4?
 

1. How does gravity affect the movement of a small lump of ice sliding down a large, smooth sphere?

Gravity plays a major role in the movement of the small lump of ice as it slides down the large, smooth sphere. As the ice moves down the slope, it is pulled towards the center of the sphere by the force of gravity. This force acts as a constant acceleration on the ice, causing it to gain speed as it slides down the sphere.

2. What factors affect the speed of the small lump of ice as it slides down the large, smooth sphere?

The speed of the ice is affected by several factors, including the slope of the sphere, the mass of the ice, and the force of gravity. A steeper slope will result in a faster slide, while a larger mass will require more force from gravity to accelerate. The smoothness of the sphere's surface can also impact the speed, as a rough surface may create more friction and slow down the ice's movement.

3. How does the shape of the small lump of ice affect its movement down the large, smooth sphere?

The shape of the ice can affect its movement down the sphere in several ways. A larger surface area may create more friction and slow down the ice's movement, while a more compact shape may allow it to slide more smoothly. The shape can also affect how the ice interacts with the sphere's surface, potentially causing it to roll or tumble as it moves down.

4. Can the temperature of the sphere or the ice impact the sliding motion?

Yes, the temperature can play a role in the sliding motion of the ice. A colder temperature can make the ice more brittle, potentially causing it to break or shatter as it slides down the sphere. A warmer temperature may also affect the friction between the ice and the sphere's surface, impacting the speed and smoothness of the slide.

5. What other factors may influence the movement of the small lump of ice on the large, smooth sphere?

Other factors that may influence the movement of the ice include external forces such as wind or other objects, the presence of any imperfections or obstacles on the sphere's surface, and the initial velocity or direction of the ice before it begins sliding. Additionally, any changes in the sphere's environment, such as a change in gravity or added moisture, may also impact the ice's movement.

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