Calculating Particle Descent on a Smooth Sphere: Practical Solution Guide

[Kushal]

circular motion please

Homework Statement

a particle is slightly displaced from rest at the top of a fixed smoothsphere of radius r. Find the the vertical height through which the particle descends befor eleaving the sphere.

Homework Equations

f = (mv^2)/r
w = mg

The Attempt at a Solution

i thought that the particle would leave the sphere when the centripetal force is less than the contact force. the contact force is equal to the x component of weight. from the angle, i found the height, i.e. cos (theta) = h/r h is the adjacent, and r is the hypotenuse.i took the angle between the negative x-axis and the actual weight vector as theta.

i can't find the answer and I'm afraid I've messed up the concepts

 

[Bryan52803]

Someone please correct me if I'm mistaken, but they way I see it, there isn't any centripetal force involved, only gravity affecting a particle on a variable sloped plane. That said, logically from the description you give me I assume that there is no friction and thus since it is jarred from rest and no longer on a flat plane it will move down the side of the sphere with an increasingly steeper slope. However, when it reaches the "equator" of the sphere, it has no force compelling it to adhere to the spheres surface. So, with simple geometry, you can find the total height it falls.

 

[Kushal]

eerrrrmmm...btw the anmswer is r/3. the particle leaves well before the equator.

 

[Astronuc]

Find the equation for the normal force of the mass on the spherical (circular) surface, which goes to zero when the particle just as the particle leaves the surface. Resolve the weight, mg, into tangential and normal components to spherical surface, and think about the expression for centripetal acceleration.

http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html

Also the particle falls a distance h, so find the kinetic energy of the particle based on the tangential velocity, and that kinetic energy must equal the change in gravitational potential energy after falling h.

 

[Kushal]

i still don't get it...

i found the normal to be -mg cos 'theta'
i took the angle to be between the weight vector and and the normal component to the circle.

i equated mg cos 'theta' to zero, but i can't seem to get anything.

i also tried equating centripetal force to the normal component...but i didn't get the answer.

 

[Astronuc]

i found the normal to be -mg cos 'theta'
i took the angle to be between the weight vector and and the normal component to the circle.

Correct

i equated mg cos 'theta' to zero, but i can't seem to get anything.

But of course the weight is not zero.

i also tried equating centripetal force to the normal component...but i didn't get the answer.

mg cos\theta - ma = 0, which indicates that the normal component holding the block to the circular surface is the gravitational component pointing toward the center of the arc, and a is the centripetal acceleration.

There is also another expression for a in terms of the tangential velocity and the radius.

Then one needs to use the conservation of energy to equate the change in kinetic energy to the change in potential energy (i.e. let's neglect friction).