What is the maximum height the pinball reaches on the inclined surface?

[whoknows123]

A ball (solid sphere of mass=0.1 kg, outer radius=0.1 m) is cocked back 0.7 m on a spring (k= 50 N/m), and on a tilted surface.

How much energy is stored in the spring initially?
(I found that to be 12.25 J)

Find the maximum height the pinball rolls to (above its initial location.)
How do I find the maximum height given only those information.

 

[Tide]

What have you tried so far?

HINT: Energy is conserved.

 

[whoknows123]

E= (1/2)mv^2 + (1/2)kx^2

 

[whoknows123]

but i don't have velocity

 

[whoknows123]

12.25 J = (1/2)(.1kg)v^2 + (1/2)(50 N/M)x^2 but i have no idea what velocity is

 

[Lee]

Exactly, but through some anaylsis of energy conservation and the above formula I'm sure you could find one.

 

[Tide]

But you do have a velocity. You know the velocity is zero just as the spring begins to accelerate the mass. And, if the object is confined to travel along the incline during acceleration then energy conservation gives you the velocity (both components!) just as the object leaves the incline.

 

[Lee]

Well what you've put above is not strictly true, when the spring is released your energy is conserved so all your potential is turned into kinetic energy...

 

[whoknows123]

anyway of helping me find velocity with those give information?

 

[whoknows123]

velocity intial is zero, but what about velocity final?

 

[whoknows123]

A = velocity of obj times squareroot(m/k)?

 

[Lee]

you were very close to the right answer with 12.5J. That is E, now when the object is being released from the sling it then has no potential energy, but has a lot of kinetic, so...

 

[Pengwuino]

Think about the situation, when it reaches the top of that incline, what's its velocity?

 

[whoknows123]

PE=KE? but KE=(1/2)mv^2 (there's no distance for me to find)

 

[whoknows123]

velocity would be zero wouldn't it?

 

[whoknows123]

nvm i take that back, i was thinking of something else

 

[whoknows123]

E = (1/2)kx^2?

 

[Lee]

Yes, that is the enregy from the sling that you worked out initially, but what does that also equal from your conservation of energy.

 

[Pengwuino]

You can simplify this by figuring out 2 things

Initial: Potential + Kinetic

You know that the velocity is 0 so its all potential.

Since you know energy is conserved, you have:

Final: Potential + Kinetic

Again the velocity would be 0 again.

Thus, there must exist a potential at the final position. What do you think it is?

 

[whoknows123]

thanks guys, i think i got it

 

[arunbg]

I think we are missing out on the rotation part.
First u sould equate the spring PE with the translational KE of the ball (assuming the spring doesn't impart rotation).Then static friction on the plane allows rolling with an associated Angular KE.
Translational KE + Angular KE = Gravitational PE at the top.
.5kx^2 + .5Iw^2 = mgh
For pure rolling,
w=v/r
I think you can go from here.

Arun

 

[Tide]

Don't forget to account for the change in gravitational potential.

 

[Pengwuino]

Im fairly sure you can ignore rotatoinal energy since it's beginning at rest and ending with 0 velocity

 

[arunbg]

Im fairly sure you can ignore rotatoinal energy since it's beginning at rest and ending with 0 velocity

Then why should you even take translational KE ?
You don't say they have given details about the wheel for nothing do you?

 

[Pengwuino]

it isn't a wheel

 

[arunbg]

Well, its has the cross section resembling a wheel at least.

 

[Curious3141]

I think we are missing out on the rotation part.
First u sould equate the spring PE with the translational KE of the ball (assuming the spring doesn't impart rotation).Then static friction on the plane allows rolling with an associated Angular KE.
Translational KE + Angular KE = Gravitational PE at the top.
.5kx^2 + .5Iw^2 = mgh
For pure rolling,
w=v/r
I think you can go from here.

Arun

I disagree. Since the final state of the ball up the incline is complete rest with zero angular velocity and there are assumed to be no dissipative forces acting upon it, we can neglect the rotational (and indeed the translational) kinetic energy components entirely.

BTW, a dissipative force is one that does work on an object and causes heat (or sound energy) loss. Kinetic friction is a dissipative force while static friction is a conservative force.

In this case, if the ball is assumed to roll without slipping, there are no dissipative forces acting on it. The static friction (which is less than or equal to the limiting friction since there is no slip) does no work on the object. Hence all of the spring's potential energy at the start goes to increasing the translational KE, rotational KE and gravitational PE of the ball as it rolls up the slope. During the roll, there is no heat or sound generation, hence mechanical energy is conserved. The sum of the individual varying KEs and PE is a constant and is equal to the elastic potential energy of the cocked spring at the start. When the ball comes to a standstill, this total is still conserved, only everything is in the form of gravitational PE. So the problem reduces to a trivial comparison of change in gravitational PE = loss in elastic energy of the spring as it uncoils.

But how do we know the ball doesn't slip? Short answer, we really don't, but we have to assume it doesn't because the question says it "rolls" but doesn't give us enough info one way or the other to determine for ourselves if the limiting friction is exceeded by the initial elastic force exerted by the spring. We have to assume that the surface is rough enough that the initial force on the ball is not greater than the limiting friction $$F_{fr} = \mu_s F_N$$. If this condition were violated, what would happen is that the ball would slip and roll simultaneously until enough dissipative work is done by the kinetic frictional force to deplete the translational KE, leaving only the rotational KE component (which would be sustained by the non-dissipative static frictional force).

So, although I've gone through some complexities in my explanation here, this problem should be assumed to be really simple, for the reasons stated.