Answer:
[tex]\theta=\cos^{-1}\dfrac{2}{3}[/tex]
Explanation:
Given that,
Mass = m
Radius = r
Angle [tex]\theta=\cos\dfrac{2}{3}[/tex]
Prove that,
The body leaves the sphere when [tex]\theta=\cos^{-1}\dfrac{2}{3}[/tex]
When the particle was contact
[tex]mg\cos\theta=\dfrac{mv^2}{R}[/tex]
[tex]Rg\cos\theta=v^2[/tex].....(I)
Using conservation of energy
[tex]mg(R-R\cos\theta)=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2gR(1-\cos\theta)[/tex]....(II)
From equation (I) and (II)
[tex]Rg\cos\theta=2gR(1-\cos\theta)[/tex]
[tex]\cos\theta=2-2\cos\theta[/tex]
[tex]\theta=\cos^{-1}\dfrac{2}{3}[/tex]
The body leaves the sphere when [tex]\theta=\cos^{-1}\dfrac{2}{3}[/tex]
Hence proved.
The angle when the body leaves the sphere is [tex]\theta = cos^{-1} (\frac{2}{3} )[/tex], proved.
The force on the particle when it is stationary is calculated as follows;
mgcosθ = mv²/R
v² = Rgcosθ ---(1)
The final speed of the object when it starts to slide can be determimed using conservation of energy.
mg(R - Rcosθ) = ¹/₂mv²
v² = 2gR(1 - cosθ) ---(2)
solve (1) and (2) together
Rgcosθ = 2gR(1 - cosθ)
cosθ = 2 - 2cosθ
3cosθ = 2
cosθ = 2/3
[tex]\theta = cos^{-1} (\frac{2}{3} )[/tex]
Thus, the angle when the body leaves the sphere is [tex]\theta = cos^{-1} (\frac{2}{3} )[/tex], proved.
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