What is the marble's speed at the top?

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Homework Help Overview

The discussion revolves around two physics problems: one involving the motion of a marble in a vertical circular path and another concerning the compression of a vertical spring when a block is released. The marble's speed at the top of the circle and the behavior of the spring under different conditions are central to the questions posed.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of energy conservation principles to find the marble's speed at different points in its path and question the setup of the spring problem, particularly regarding the differences in compression based on how the block is released.

Discussion Status

Some participants have provided guidance on drawing angles for the marble problem and clarified the relationship between force and compression for the spring. However, there are still questions about the assumptions made regarding the marble's motion and the effects of rotational inertia, indicating ongoing exploration of the concepts.

Contextual Notes

Participants note that additional information, such as the mass and size of the marble, may be necessary to fully address the rotational aspects of the marble's motion. There is also a distinction being made between the scenarios of dropping the block versus lowering it slowly, which affects the spring's compression.

kw_cecilia
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I am doing these questions, and I don't quite understand.

1. A marble spins in a vertical plane around the inside of a smooth, 20cm-diameter horizontal pipe. The marble's speed at the bottom of the circle is 3.0m/s

a) what is the marble's speed at the top?
b) Find an algebraic expression for hte marble's speed when it is at angle phata, where the angle is measured from the bottom of the circle.

For part a, I used

1/2mvf^2 + mgyf = 1/2 mvi^2 + mgyi

to find vf , and i got 2.25 m/s

But for part b, i don't understand where should I draw the angle.

2. A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touching it.

a) How far does the spring compress if you let go of the block suddendly?
b) How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?
c) Why are your two answers different?

How do I deal with questions with a vertical spring?
 
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Hi,

If I am reading you question right...

For the fist part:

Draw a circle then draw a vertical radius from the centre to the base. Now draw any radius from the centre to another part of the circumference. The angle between the two is the angle required.

Use the same equations as before, but use trig to find the the height.

For the second part:

his spring will compress rather than extend. The equations/method is exactly the same as for extension springs but now x as in "F=kx" is the compression from the natural length, not the extension!

Hope that helps

EaGG
 
Oo.. thankyou! Now I understand the first part !
But for the second part, does it mean the mg of block = kx?
 
Sorry not to get back...bit busy!

mg = kx ...yes for lowering slowly..but for the "dropping" part consider the energies involved at the top and bottom.

EaGG
 
Sorry, I didn't quite understand the answer for -1-. Does it take into account the rotational inertia of the marble? They would need to give us the mass and size of the marble in order for that to be taken into account, though.
 
I don't believe that the marble is spinning about a diametrical axis. The pipe is smooth and there is no friction involved. It's just "sliding".
 

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