- #1
JulienB
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- 12
Homework Statement
Hi everybody! Here is a classical mechanics problem about a ball (mass m, radius r) rolling down a slope (from height h) and going through a loop-the-loop.
a) What is the minimum height h from which the ball must roll in order to successfully complete the loop-the-loop without losing contact with the ramp? (Consider the radius R of the looping to be much bigger than r)
b) If the ball is rolling from a height h = 6R, what are the horizontal force components at the point Q? (see picture)
Homework Equations
Conservation of energy, centripetal acceleration, moment of inertia of a sphere (Isphere = 2/5⋅m⋅r2).
The Attempt at a Solution
a) Okay so I think I should first find the tangential velocity of the sphere as it crosses the highest point of the loop:
m⋅g = m⋅ac = m⋅(vT2/R) ⇒ vT = √(gR)
(Note: I selected R as the radius and not (R - r) because we've been told that R>>r in the problem's statement)
Then I set up an equation of conservation of energy:
PE0 = PE1 + KE1 + Erot1
⇔ m⋅g⋅h = m⋅g⋅2R + ½⋅m⋅vT2 + ½⋅Isphere⋅ω2
⇔ m⋅g⋅h = m⋅g⋅2R + ½⋅m⋅g⋅R + ½⋅(2/5⋅m⋅r2)⋅(vT2/R2)
⇔ g⋅h = g⋅2R + ½⋅g⋅R + 2/10⋅r2⋅g⋅R/R2
⇒ h = 5/2⋅R + 1/5⋅r2/R
Does that seem correct to you? (Note: again, I omitted a few "r" because of the indication R>>r)
b) Now h = 6R
I'm always a little confused about how to describe forces when it comes to centripetal and centrifugal forces, but I give it a go:
At the point Q I would say that the horizontal force components are the centrifugal force FZ (that's how we usually denote it in Germany), directing towards the outside of the loop-the-loop, and the normal force (which is also the centripetal force in that case), directing towards the center of the loop-the-loop.
Because there is no horizontal acceleration at that point Q (the velocity of the ball points upwards), I would assume that FZ = FN = m⋅ac = m⋅vT2/R
To find the tangential velocity at that point, I set up my equation of conservation of energy again but solving this time for vT2 and with h = 6R. I get vT2 = (10⋅R3⋅g)/(R2 + r2).
I substitute then that tangential velocity into my first equation and get:
FN = FZ = (10⋅m⋅R3⋅g)/(R2 + r2)
Is that correct? I'm preparing an exam, so I'd like to make sure I am able to solve such problems, and if not hopefully I can correct the mistakes I still make.Thank you very much in advance for your answers, I appreciate it.Julien.