What Force Acts on a Marble at Point Q in a Loop-the-Loop?

[slk011]

Help please -- marble rolling around loop-the-loop

A solid marble of mass m = 15 kg and radius r = 8 cm will roll without slipping along the loop-the-loop track shown in the figure if it is released from rest somewhere on the straight section of track. The radius of the loop-the-loop is R = 1.20 m.


1- (which i got) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop?
3.24 m

2- If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?


SOOO lost in this one!


Diagram http://www.webassign.net/hrw/hrw7_11-32.gif

 

[azizlwl]

Maybe you can show how you get the answer to the first question.

 

[Rayquesto]

Well, when this marble circulates, you would suspect that a normal force and a centripetal force as well as weight would exist to keep the marble in circular motion, however, at that Q point, only centripetal force and normal force exists in the x direction.

 

[dipole]

Well, when this marble circulates, you would suspect that a normal force and a centripetal force would exist to keep the marble in circular motion, however, at that Q point, only centripetal force exists. Therefore, Q does not exert a force on the marble, because no normal force exists. What do you think?

I don't think so. The normal force IS what is providing the centripetal force, and it is what is contributing to the horizontal component.

Use the expression for circular motion to solve for this component, don't forget to include the effect of gravity.

 

[Rayquesto]

Hm... I see what you are saying. The only reason why I brought up the idea was because I thought of weight; It only has a y component. Thanks!

 

[collinsmark]

A solid marble of mass m = 15 kg and radius r = 8 cm will roll without slipping along the loop-the-loop track shown in the figure if it is released from rest somewhere on the straight section of track. The radius of the loop-the-loop is R = 1.20 m. 1- (which i got) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop?
3.24 m

You answer is close, but not quite right. You handled the translational and rotational kinetic energy concepts fine, but there is a detail you must have left out. Let me give you some hints:

When the ball is at the top of the loop, is its center of mass at a height of 2R, or is it less than that? (Remember, the ball has its own radius of r.)

When the ball is circling in the loop, is the ball circling with a radius (according to the ball's center of mass) of R, or a little less than that? (Once again remember, the ball has its own radius of r.)

[Edit: this detail will have a small, yet significant impact on the final answer.]

2- If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?

Use conservation of energy like you did in the first part. However this time, instead of solving for h, you need to solve for the ball's linear speed, v, at point Q. Then find the centripetal force.

 

[slk011]

Collinsmark was my 3.24m. Is what you are referring as not right?
Would v=sqroot g(R-r). Be the right track?

 

[collinsmark]

Collinsmark was my 3.24m. Is what you are referring as not right?
Would v=sqroot g(R-r). Be the right track?

Yes. In terms of the radius of curvature of the ball orbiting on the loop, the centripetal acceleration's magnitude is
[tex] a = \frac{v^2}{R-r}. [/tex]
(Substitute g for a if you want the minimum speed that keeps the ball from leaving the track at the top of the loop.)

But that's not all. You must also consider the height of the ball above the ground when it is at the top of the loop. And that height is dependent, in part, upon the ball's radius [Edit: this is only necessary for Part 1 of the problem]. This is part of the equation which represents the difference in gravitational potential energy.

Taking the radius of the ball into account won't change your final answer by a whole lot, but it will change it by a non-negligible amount. It's significant enough to take into consideration.

 

[slk011]

It said 3.34 was right tho.. :(. I still don't understand how to get part B

 

[collinsmark]

It said 3.34 was right tho.. :(.

If R is measured from the center of the loop-the-loop to the edge of the track, 3.24 meters is close, but not quite right (I assume that "3.34" is a typo, and you meant 3.24. [STRIKE]3.34 meters isn't correct either)[/STRIKE].

[Edit: I take that back, 3.34 meters is very close. I just want to make sure you know where the ~3.34 meters comes from, and why it's not 3.24 meters when you consider the ball's radius. In the original post, you had calculated 3.24 meters.]

On the other hand, if R is the distance from the center of the loop-the-loop to the center of the ball, then 3.24 meters is correct for part 1. But that's not the way that I interpreted the problem statement, but maybe that's what you were supposed to assume anyway (could have been a poorly constructed problem statement or problem).

I still don't understand how to get part B

Use essentially the same approach as you did before: Conservation of energy. What is the gravitational potential energy difference from when the ball is at height 6R compared to when the [center of the] ball is at the height of Q? This gets converted to the ball's kinetic energy (some of that energy is rotational, some of it translational, but you already know the relationship between these, because you needed to figure that out for part 1). Solve for the ball's linear velocity (speed). Calculate the centripetal force.

 

[collinsmark]

In case you didn't see my edit, I just wanted to point out an edit I made to my last post. 3.34 m is actually quite close to the exact result, when considering the ball's radius. It's correct to within three significant figures anyway. But it's not the same as the 3.24 m figure given in the original post, is my point.