What Is the Angle When an Object Loses Contact with a Sphere?

  • Thread starter Thread starter mistymoon_38
  • Start date Start date
  • Tags Tags
    Rolling Sphere
AI Thread Summary
The discussion focuses on determining the angle at which a small object loses contact with a frictionless solid sphere as it rolls down. The key point is that at the moment of losing contact, the centripetal force equals the gravitational force acting on the object. Participants suggest using energy conservation principles to relate potential and kinetic energy, leading to the equation v^2/R = gsin(theta). By solving for the velocity and equating the two expressions, the angle of 41.8 degrees is derived. This approach effectively combines concepts of forces and energy to solve the problem.
mistymoon_38
Messages
18
Reaction score
0
Object rolling off a sphere...

Homework Statement



A small object begins at the top of a frictionless solid sphere. Its initial speed is negligibly small. The sphere is stationary at all times. The small object then slides down the surface of the sphere. At one point the small object loses contact with the sphere. Draw a line from this point to the center of the sphere. What is the angle between this line and the horizontal?

Homework Equations





The Attempt at a Solution



I have no idea how to get this started. Any hints on where to go with this?
 
Physics news on Phys.org
Whenever something loses contact with a surface the normal force vanishes.
 


Hey,

I bet you're in my recitation unless our professors somehow took the same problem! Although this problem requires the use of energy, we need to understand that the centripetal force is equal to the gravitational force at the point when the object leaves the sphere.

This gives us v^2/R = gsin(theta). Solve for the velocity here.

Then you want to use the Conservation of Energy equation. You know that the initial potential energy is mgR (we assume the potential energy line is a horizontal line through the center of the sphere). Then when you solve for "Mechanical Energy Final", we know that the potential energy is smaller (mgRsin(theta)) and the kinetic energy is 1/2mv^2. I recommend you solve for v in the energy equation and then solve the two v's against each other.

Through this process, you should find the angle. I got 41.8 degrees, let me know if you disagree.
 
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