Centripetal force motorcycle problem

AI Thread Summary
The discussion revolves around determining the maximum speed of a motorcycle traveling over the crest of a hill with a circular arc radius of 45 meters without losing contact with the road. Key concepts include understanding the forces acting on the motorcycle, specifically gravitational force (mg) and the normal force (FN) at the top of the hill. The confusion arises from the assumption that FN equals mg; however, at the point of losing contact, FN must actually be zero. The correct approach involves recognizing that the net force must provide the necessary centripetal acceleration, leading to the conclusion that FN should be zero when the motorcycle is at the crest. This clarification helps in solving the problem accurately.
bulbasaur88
Messages
57
Reaction score
0
A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of 45.0 meters. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

Will somebody sound this problem out with me please? I think I am missing an important concept that I am missing...

Given
r = 45 m
Vmax = ____ m/s

What I Gather
We do not want the motorcycle to lose contact with the road, especially at the zenith.

At the very top of the hill, we have mg pointing down towards the center of the circle. We also have FN, which points up and away from the center of the circle.

***I hope the above forces have been set up correctly in my free body diagram.***

My question is: Doesn't FN = mg ?

That would give me: FC = mv2 / r = mg - FN = 0

? I get lost here...I am sure there is something wrong with either my FBD or my reasoning. Can I get a hint please?
 
Physics news on Phys.org
bulbasaur88 said:
At the very top of the hill, we have mg pointing down towards the center of the circle. We also have FN, which points up and away from the center of the circle.

***I hope the above forces have been set up correctly in my free body diagram.***
Looks good.

My question is: Doesn't FN = mg ?
No. If FN = mg, then the net force on the motorcycle would be zero--no acceleration at all.

Hint: What must FN be when the motorcycle just loses contact with the road?
 
Zero! - thank you :))))
 
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