A marble rolls down a track, the bottom of which bends into a loop. The top of the track is 0.8m above the ground. Point B is the highest point of the loop. The diameter of the loop is 0.35m. The mass of the marble is 0.035 kg and gravity is 10 m/s². What is the speed of the marble at point B?

Question

deoxna · Answer

|dw:1330306581225:dw| This is a diagram of the problem

anonymous · Answer

Humm. In order to figure out the speed of the marble when it hits the loop we need an angle or the length of one of the other sides of the triangle...

anonymous · Answer

Is it assumed that the length and height of the triangle are equal?

deoxna · Answer

Sorry, I think the marble might have been misleading. The marble is dropped at point A, which is 0.8 meters from the ground, The ground is also tangent to the loop, so I guess a point C at the bottom of the loop is directly opposite to B. Hope that helps. I haven't practiced with circular motion much so I'm not really sure how to adress this problem.

anonymous · Answer

You still need another length or an angle so you can determine the speed of the marble when it hits the start of the loop. Gravity would then affect the velocity and speed at the top of the loop could be determined.

deoxna · Answer

The only other information the question gives is that we must "dedice that the speed of the marble is 3 m/s²" but since that essentially means to show work, it doesn't get us any closer to the process.

deoxna · Answer

Couldn't change in potential energy help somehow?

anonymous · Answer

At what point is the marble going 3m/s? At the top of the loop? And you have it as 3m/s2 which would be an acceleration. Is that correct?

anonymous · Answer

Use conservation of energy: 1/2mv1^2 + mgh1 = 1/mv2^2 + mgh2
I got: V2 = 3 m/s

anonymous · Answer

Oops.  The second kinetic energy is supposed to be: 1/2mv2^2