What is the marble's speed at the top?

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SUMMARY

The discussion focuses on calculating the speed of a marble at the top of a vertical circle and the behavior of a spring when a block is released. For the marble, using the conservation of energy equation, the speed at the top is determined to be 2.25 m/s. The second part involves a vertical spring with a spring constant (k) of 490 N/m and a 5.0 kg block, where the compression of the spring is analyzed under two scenarios: dropping the block and lowering it slowly. The key takeaway is that the equations for spring compression apply similarly to both scenarios, but the energy considerations differ.

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  • Understanding of conservation of energy principles in physics
  • Familiarity with spring mechanics, specifically Hooke's Law (F = kx)
  • Basic knowledge of trigonometry for angle calculations
  • Concept of rotational inertia and its relevance in circular motion
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  • Study the conservation of energy in circular motion, particularly for objects in vertical circles
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Students and educators in physics, particularly those studying mechanics, energy conservation, and spring dynamics. This discussion is beneficial for anyone looking to deepen their understanding of forces and motion in practical applications.

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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