Ask Question
3 April, 14:30

An object is at rest on top of a smooth sphere with a radius of? = 15.3 m that is buried exactly halfway under the ground. It then begins to slide down. At what height from the ground is the object no longer in contact with the sphere?

+4
Answers (1)
  1. 3 April, 14:39
    0
    10.2 m.

    Explanation:

    Let object falls by angle θ.

    At any moment after the fall, there are two forces acting on the sphere

    1) mg cosθ which is a component of weight towards the centre 2) normal reaction of the surface R.

    mgcosθ - R is net force acting, which provides centripetal force

    mgcosθ - R = mv² / r

    But v² = 2g r (1-cosθ) [ object falls by height (r - r cosθ).

    mgcosθ - R = m / r x 2g r (1-cosθ)

    When the object is no longer in touch with sphere,

    R = 0

    mgcosθ = m / r x 2g r (1-cosθ)

    3 gr cosθ = 2gr

    cosθ = 2/3

    height of fall

    = r (1-cosθ)

    r (1 - 2/3)

    1/3 r

    1/3 x 15.3

    5.1 m

    Height from the ground

    15.3 - 5.1

    10.2 m.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “An object is at rest on top of a smooth sphere with a radius of? = 15.3 m that is buried exactly halfway under the ground. It then begins ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers