Post a New Question


posted by .

A particle of mass m is released from rest at the top of a spherical dome of radius R.
how far below the starting point will the particle leave the surface of the dome?
how i should solve this?

  • physics -

    They need to say whether the particle slides or rolls. If it slides, you have to know the friction coefficient. If it rolls, you have to know if it is a solid or hollow sphere.

    If it slides without friction, we can do the problem. After descending a vertical distance H, it will have acquired a speed
    V = sqrt(2gH)

    Let A be the angle than it has cescended, measured from the center of the sphere

    It leaves the sphere when the componemt of its weight normal to the sphere, M g cos A, is equal to the centripetal force required to make it follow the circular trajectory. When this happens, the sphere no longer needs to apply a reaction force to the particle to keep it there, and the particle leaves the surface.

    So require that
    M g cos A = M V^2/R = M *2g H/R

    cos A = 2H/R

    Geometry also tells you that
    H = R (1-cos A)

    cos A = (1-cos A)
    cos A = 1/2

    A = 60 degrees

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question