Pinball machine launch ramp consisting of a spring

shootmewithproton
Posts: 4
Joined: Sun Jun 22, 2014 6:20 pm

Pinball machine launch ramp consisting of a spring

Postby shootmewithproton » Thu Aug 07, 2014 10:11 pm

Hi. I am working out the problems on Conquering the GRE and here is the question I got stuck:

Question:

A pinball machine launch ramp consisting of a spring of force constant k and a 30 degree ramp of Length L. What is the ball's speed immediately after being launched? The ball has m mass and r radius.

Attempt:

Well, to begin with, this is a 3 question problem and this is the second question. From the first question, I could obtain how much the spring needs to be compressed just to make the top of the ramp without rolling back (and friction is sufficient that the ball begins rolling without slipping after lunch).

so compressed distance x is sqrt(m*g*L/k) which I confirm to be correct.

We wanna know the speed of the ball immediately after being launched. So set up the conservation of energy equation such that

0.5*m*v^2 = 0.5*k*x^2 since all the potential energy is converted into kinetic energy.

Simplify it to m*v^2 = k*x^2. We know x. so plug x in.

k*m*g*L/k = m*g*L which is equal to m*v^2.

Hence, v = sqrt(g*L).

And the book says it is wrong with other argument explaining with accounting both transnational rotational energies... I understand their calculation but what I don't understand is where my argument went wrong about setting all potential energy being converted into kinetic.

Can you please help me out?

TakeruK
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Joined: Mon Jan 02, 2012 3:05 pm

Re: Pinball machine launch ramp consisting of a spring

Postby TakeruK » Sat Aug 09, 2014 2:25 pm

I think you computed the potential energy correct, but as the book's solution says, the total kinetic energy has two parts: translational and rotational. There is rotational kinetic energy because the pinball is rolling without slipping, not just sliding along the surface.

blighter
Posts: 256
Joined: Thu Jan 26, 2012 6:30 pm

Re: Pinball machine launch ramp consisting of a spring

Postby blighter » Sun Aug 10, 2014 2:38 pm

You better ask Tommy. : )

shootmewithproton
Posts: 4
Joined: Sun Jun 22, 2014 6:20 pm

Re: Pinball machine launch ramp consisting of a spring

Postby shootmewithproton » Mon Aug 11, 2014 5:18 pm

@TakeruK

But we are asked to compute the ball's speed immediately right after it is being launched. So why do we have to consider rotational and kinetic energies?

Oh, hold on, so we are assuming that as soon as the ball is launched, the friction kicks in right away, thus we have to account the rotational energy too?

If that is the case, then it makes sense to me (meaning my energy conservation set up is just off by rotational energy portion).

TakeruK
Posts: 812
Joined: Mon Jan 02, 2012 3:05 pm

Re: Pinball machine launch ramp consisting of a spring

Postby TakeruK » Mon Aug 11, 2014 9:47 pm

You are computing the K.E. just before the ball starts to both move and roll at the same time, so you need to account for both energies, like you said.

cgmusselman
Posts: 1
Joined: Thu Aug 20, 2015 11:42 am

Re: Pinball machine launch ramp consisting of a spring

Postby cgmusselman » Thu Aug 20, 2015 11:53 am

I need to know how far the plunger was pulled (= distance spring stretched) to know how much energy input.

Chiron
Posts: 9
Joined: Tue Sep 01, 2015 10:37 am

Re: Pinball machine launch ramp consisting of a spring

Postby Chiron » Tue Sep 01, 2015 1:24 pm

This is Q3 - Q5 from section 1.3.5 in "Conquering the Physics GRE".

For Q4, the one which asks about the ball's speed immediately after being launched, I think that when it says immediately after being launched that it means just as the ball is leaving the spring. Thus, the ball itself has already moved a non-zero distance by that point, and is thus rolling. Thus, your kinetic energy equation should include \dfrac{1}{2} m v^2 + \dfrac{1}{2} I \omega^2.




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