Loop the loop problem with centripetal force

AI Thread Summary
The discussion revolves around solving a physics problem involving a block sliding on a loop-the-loop track. For part (a), the centripetal force at point Q can be determined by first calculating the block's speed using conservation of energy principles. In part (b), a free body diagram is essential to analyze the forces acting on the block at the top of the loop to find the height from which it should be released. Part (c) involves comparing the forces at the top and bottom of the loop, which can be derived from the previous calculations. The conversation emphasizes the importance of applying both energy conservation and centripetal force equations to solve the problem effectively.
r19ecua
Messages
17
Reaction score
0

Homework Statement


A small block of mass m = 5kg slides along the frictionless loop-the-loop track shown in the figure. R = 2m

a) If it starts from rest at P, what is the resultant centripetal force acting on it at Q?
b) At what height from the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight?
c) What is the difference between the force at the top and the bottom?

Homework Equations


m*v^2/r
1/2mv^2f + mgyf = 1/2mv^2i + mgyi
This is as far as I know..

The Attempt at a Solution



Well I thought this was a simple conservation of energy problem, and then it HAD to involve centripetal force. Now I have no idea how to approach this problem :( This is a review problem, please help!
 
Physics news on Phys.org
For part (a), using your conservation of energy formula will you the speed of the block at point Q. Once you get the speed at that point, your first equation gives you the resultant centripetal force at that point. So you have the correct relevant equations needed to solve part (a).

For part (b), draw a free body diagram at the point in question, and note all the forces acting on it. The sum of those forces in the centripetal direction is the net centripetal force in that direction. Then go back to your energy equation to solve. The solution to part c should follow.

I assume the figure which is not attached gives other numerical values so that you can numerically solve the problem.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top