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stealth61 [152]
3 years ago
10

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. I

t then begins to slide down. At what height from the ground is the object no longer in contact with the sphere?
Physics
1 answer:
NeX [460]3 years ago
4 0

Answer:

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 .

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q_1 = -1 \cdot 10^{-6}C
q_2 = +3 \cdot 10^{-6 } C
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