Can a Ball Roll in a Horizontal Circle Inside a Bowl?

AI Thread Summary
A ball can roll in a horizontal circle inside a smooth bowl if specific conditions regarding its speed and height are met, as described by the equation v^2 = gh(2a-h/a-h). The discussion emphasizes the importance of understanding centripetal acceleration and the forces acting on the ball, particularly in a frictionless environment. The role of gravitational and centrifugal accelerations is highlighted, suggesting that for stable circular motion, the acceleration vector must align with the bowl's curvature. Drawing free-body diagrams is essential for visualizing the forces and ensuring the ball remains at a constant height. Understanding these principles is crucial for solving the circular motion problem effectively.
Nick886
Messages
2
Reaction score
0
circular motion problem??

hey all, i have to talke app maths for a semester and it was all fine until we got to modeling, now haven't a clue? can anyone help with this please?
a ball rolls under gravity inside a smooth bowl with circular cross section and inner radius"a". show that the ball can roll in a horizontal circle at a height "h" above the bottom of the bowl provided
v^2 = gh(2a-h/a-h). thanks
 
Physics news on Phys.org


What do you know about circular motion? Can you calculate centripetal acceleration?

Have you drawn a diagram of the setup, and a free-body diagram for the forces acting?
 


ye i have done that, well tried! there is no friction because its a smooth bowl. so is it just gravity or am i missing something obvious?? Probably! i got as far as the cent acc and have \omega^2 = g(tan\vartheta / a)
 


Hint: Consider for a moment, in an appropriate frame of reference where such things are not frowned upon, the sum of the centrifugal acceleration and the gravitational acceleration that the ball feels. The bowl's reaction is equal and opposite, so draw in the reaction acceleration. Here's the tricky bit. If this vector is not aligned with the surface normal of the bowl, that is, aligned with the radius vector of the bowl (curvature radius a) at the point of contact, then there will be a net acceleration away from the point of contact, either upward or downward, tending to shift the height of the ball.

If this acceleration vector must be aligned with the radius vector for stability of the rotation height, then a simple ratio follows.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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