How Much Should the Spring Be Compressed to Keep the Mass on Track?

[Jamest39]

Homework Statement

A 50-gram mass is accelerated from rest be a compressed spring (k = 1800 N/m), sending it on a journey along a frictionless loop-de-loop of radius 0.324 m. What minimum amount of initial compression of the spring is required if the mass is to remain in contact with the track at the top of the loop?

Homework Equations

centripetal acceleration = (v^2)/r
F = ma
F = kx (Hooke's Law)

The Attempt at a Solution

The velocity required at the top of the loop:
v = sqrt( 9.80m/s^2 * 0.324 m) = 1.782 m/s
Since the spring constant k is already known, we'd just need to determine F and use Hooke's Law to solve for x. But I can't figure out how to determine the force required to make push that mass and give it that velocity at the top of the loop.

 

[steven george]

I would think about using the potential energy of the spring, 1/2 kx². That amount should be equal to the kinetic energy for the speed you need added to the gravitational potential energy for a height equal to the diameter of the loop.

 

[Jamest39]

I would think about using the potential energy of the spring, 1/2 kx². That amount should be equal to the kinetic energy for the speed you need added to the gravitational potential energy for a height equal to the diameter of the loop.

So the gravitational potential energy is just mgh = (0.05 kg)(9.8 m/s^2)(0.648m)
But how can I solve for the potential energy of the spring without x (the amount of compression)?

 

[The Vinh]

I have looked through your OP and I will show you how to find X. But before I tell you, have you learned popential energy and kinetic energy ? I don't want to write down an answer without your understanding.

 

[The Vinh]

And does your homework provide the distance between the spring and the loop - de - loop ( sound like french)

 

[Jamest39]

And does your homework provide the distance between the spring and the loop - de - loop ( sound like french)

yes, we learned about it, I have the formulas for kinetic and potential energy. And it does not provide any distance, so I guess we can assume it starts right on the circle?

 

[The Vinh]

Sound good, in order to let the substance remain contact with the track at the top of loop - de - loop, centripetal acceleration must be greater than gravity, Ac >= G. It means that V^2 = gr.
+) Frame of reference is the ground
+) 1/2V^2 + mg2r = 1/2Vo^2 => Vo^2 = gr + 2g2r
+) you use the conservation of energy: 1/2kx^2 = 1/2mVo^2 => you can find X ( i have to let you do the rest )

 

[The Vinh]

For the spring, i choose the original length as frame of reference

 

[Jamest39]

For the spring, i choose the original length as frame of reference

So then, using those, we would get:
x = sqrt( (10mgr)/k )

 

[The Vinh]

So then, using those, we would get:
x = sqrt( (10mgr)/k )

That's right. And are you learning Pascal programming language ? because i saw sqrt()

 

[The Vinh]

Oh i am sorry, your answer wasn't right. x = sqrt((5mgr)/k)

 

[Jamest39]

Oh i am sorry, your answer wasn't right. x = sqrt((5mgr)/k)

Ah, I see. I dropped one of the 1/2 while working through it. Thanks!
And I'm not learning that programming language, that's just how I have to enter in square roots on my online math homework.

 

[The Vinh]

Ah, I see. I dropped one of the 1/2 while working through it. Thanks!
And I'm not learning that programming language, that's just how I have to enter in square roots on my online math homework.

I see

 

[steven george]

So the gravitational potential energy is just mgh = (0.05 kg)(9.8 m/s^2)(0.648m)
But how can I solve for the potential energy of the spring without x (the amount of compression)?

If you know how much potential energy it has at the top, as you have described then just add the kinetic energy for the speed that you have already calculated. That will give you the total amount of energy. All of that energy comes from the potential energy of the spring so the total energy is equal to the elastic potential energy and "x" is the only thing that you don't know in the equation.