College Physics problem! Anyone know the angle at which a mass sliding off a circular object goes airborne?

Question

It is supposed to be a uniform angle, not dependant on mass, weight, velocity etc. The mass starts at the top of the circular object and begins sliding down the circle eventually there is a point when it gets airborne. Anyone know?? I have tried searching and can't find answer! Thanks

It is supposed to be a uniform angle, not dependant on mass, weight, velocity etc. The mass starts at the top of the circular object and begins sliding down the circle eventually there is a point when it gets airborne. My professor said it is 48 degrees but I can't figure out how. I have tried searching and can't find answer! Thanks

Answer 1

You need to look at the normal acceleration versus the acceleration due to gravity at that angle.

The normal acceleration = v^2/R (where R is the radius and v is the velocity) and if this is greater than the component of gravity pointed towards the center of circle then airborne. So as it slides down the circle the velocity, v, is increasing, at the same time the component of gravity gets smaller.

At 0 degrees (on top) v is zero but all the gravity is pointed towards the center of the circle. At 90 degrees the velocity is equal to square root of (2 * g * R) but the gravity component passing through the center of circle is 0. So somewhere between 0 and 90 degrees it takes off.

I will let you do the math, but I would check 48 degrees as your instructor implied. (wink wink), use conservation of energy to find v.

Answer 2

something is unquestionably incorrect. right here is why: You ask a query "How a techniques will the penny fall far flung from the factor of touch of the sector and the platform?". next you're saying: "in accordance to my professor, the formulation is V^2= rgcos(theta)" this is the formulation for what??? easily no longer the formulation for the area, and that may no longer the formulation for the cost of the penny while it leaves the sector. Double examine the situation and re submit. playstation The penny will leave the exterior of the sector on the attitude theta = arc cos (2/3)

Answer 3

it's 60 degrees.
i'm not telling you whether it's from the top or the bottom or what, because you should try to do it yourself.

I'll give you several hints to get you on your way:
1) what force is required to hold you onto the circular object as your slide around it? (i.e. make you move in a circle)
2) what force is holding you ONTO the surface?

Answer 4

Yup, 90 degrees. The mass will slide round untill it's falling vertically, then it will part company with the circle and fall vertically.

Answer 5

90 degrees right 270 degrees left