Question

a small mass m starts from rest at top and slides down the smooth surface of a solid sphere of radius R. The angle of radius vector of mass with respect to vertical when it leaves the surface of sphere is ready​

Answer 1

Answer:

Angle of radius vector = 42 degree

Angle of radius vector = 42 degreeExplanation:

• Allow the lock to exit the sphere when it is at an angle θ to the vertical.
• At the point where the block exits the sphere, the sphere's typical reaction to the block is zero.
• The centripetal force is calculated by dividing the block's weight,mgcosθ, into two components.

Thus, mgcosθ = ma = mv²/R

⇒cosθ= a/g=v²/Rg

By matching the block's potential energy at the sphere's top to the KE at the point when it exits the sphere:

1/2mv²=mgh

or, v²=2gh

∴cosθ=2gh/Rg=2h/R

From geometry,cosθ=R-h/R

therefore,R-h/R=2h/Ri

mplies,

h=R/3

⇒cosθ= 2/3 =cos42∘

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