Pinball machine launch ramp consisting of a spring

 Posts: 4
 Joined: Sun Jun 22, 2014 6:20 pm
Pinball machine launch ramp consisting of a spring
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?
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?
Re: Pinball machine launch ramp consisting of a spring
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.
Re: Pinball machine launch ramp consisting of a spring
You better ask Tommy. : )

 Posts: 4
 Joined: Sun Jun 22, 2014 6:20 pm
Re: Pinball machine launch ramp consisting of a spring
@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).
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).
Re: Pinball machine launch ramp consisting of a spring
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.

 Posts: 1
 Joined: Thu Aug 20, 2015 11:42 am
Re: Pinball machine launch ramp consisting of a spring
I need to know how far the plunger was pulled (= distance spring stretched) to know how much energy input.
Re: Pinball machine launch ramp consisting of a spring
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 nonzero distance by that point, and is thus rolling. Thus, your kinetic energy equation should include $$\dfrac{1}{2} m v^2 + \dfrac{1}{2} I \omega^2$$.
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 nonzero distance by that point, and is thus rolling. Thus, your kinetic energy equation should include $$\dfrac{1}{2} m v^2 + \dfrac{1}{2} I \omega^2$$.

 Posts: 2
 Joined: Sun Aug 06, 2017 2:05 am
Re: Pinball machine launch ramp consisting of a spring
I am confused about the direction of friction force acting on the ball? I mean if there was no friction the linear velocity would be greater (squrrt(gL)) than the linear velocity when there is friction (squrrt(5gL/7)). So the downward deceleration is also less than gcos60 (no friction) when there is friction. So is the frictional force acting upward?