Quick Physics Help Please
Hey Basil,
I've been at this problem for quite a while now and I can't seem to figure out how they got to the answer. Here is the problem:
In the figure, a small block of mass m = 0.014 kg can slide along the frictionless loop-the-loop, with loop radius R = 13
cm. The block is released from rest at point P, at height h = 5R above the bottom of the loop. What are the magnitudes
of (a) the horizontal component and (b) the vertical component of the net force acting on the block at point Q? (c) At
what height h should the block be released from rest so that it is on the verge of losing contact with the track at the top
of the loop? (On the verge of losing contact means that the normal force on the block from the track has just then
become zero).
I figured out parts b) and c) but I'm confused on part a).
Diagram: http://imgur.com/qtmlXEl
It says the answer is a) 1.0976 N, but I'm unsure of how they got to that. Can someone show me the steps to it please? I've been literally at this for a few hours now, thanks so much!
5 Comments • Newest first
[quote=cb000]You can find the velocity by using conservation of energy. You only have gravitational potential initially, so E0=mgh0.
Then at point Q you have Ef=mghf+p^2/(2m)=E0. Then you can solve for the momentum.[/quote]
Confirmed but since its essentially just freefall with an upwards loop you could also do
Vf^2 = 2(9.8)(0.52) // 0.52 being 5R - R being the distance the object falls to the reach the height on the left side of the loop. Since energy is conserved throughout the system, velocity will be the 'same' on the right side.
So Vf = 3.19249119 m/s
Fnet = (0.014)(3.19249119)^2 / (0.13) = 1.0976 N as well. Should know both ways to solve it though since profs are picky but l prefer his way.
a) At Q, the block has force mg (down - vertical)
b) mv^2/R radially inward (horizontal).
c) The angular force is in the same direction as gravity at the top of the loop, so solve for mg = v^2/R
This is done using energy, dropped from point h gives it U = mgh and T = 1/2mv^2 = 0.
At the top of the loop, mgh - 2mgR = 1/2mv^2, solve for v. v^2 = 2g(h-2R). so mg = 2g(h-2R)/R, plug and chug for h.
Wrote this real quick, might have made some mistakes and I apologize if so.
You can find the velocity by using conservation of energy. You only have gravitational potential initially, so E0=mgh0.
Then at point Q you have Ef=mghf+p^2/(2m)=E0. Then you can solve for the momentum.
The centripetal formula is F=mv^2/r I believe, but I don't know what to do from here. Basically I think I would need to find velocity because the normal force would be equal to the horizontal component which is what I'm trying to get.
Wasn't there some centripetal formula involving sine and cosine.
My guess would be combining that with the total energy and using the angle of Pi/4.