Circular Motion on a Hemisphere: Finding Speed from Geometric Conditions

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SUMMARY

The discussion focuses on calculating the speed of an object in uniform circular motion on a hemisphere with a radius of r, positioned at a height of r/5 from the lowest point. The key equations used include ƩF = mv²/r and Ncosθ = mg. The correct approach involves determining the angle θ using geometric conditions, leading to the conclusion that the speed v is given by the formula v = (3√5gr)/10. The initial miscalculation of tanθ was corrected to 3/4, demonstrating the importance of accurate geometric interpretation.

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Homework Statement


As shown in attached figure, a small object is in uniform circular motion in a horizontal plane, on the smooth of a hemisphere (radius:r). The distance between the object's plane of motion and the hemisphere's lowest point is \frac{r}{5}

What is the speed of the object?


Homework Equations


ƩF=\frac{mv<sup>2</sup>}{r} ...(1)

Ncosθ=mg ...(2)



The Attempt at a Solution



From (1)
and I get ƩF from FBD wrote in attached figure ƩF=Nsinθ

From (2) I knew that N=\frac{mg}{cosθ} ..(3)

substitute (3) in (1) in got gtanθ=\frac{v<sup>2</sup>}{r} ..(4)

and I try to find tanθ from geometric of hemisphere

First, I try to find the radius (let it is r') of this mass at r/5 from the lowest point of hemisphere

If I look in the picture and use pythagoras r' = (r2-(\frac{4r}{5}))1/2
∴r' = \frac{3r}{5}
Thus; tanθ = 3
substitute in (4) v = √3gr

but the answer is \frac{3√5gr}{10}...

Or I get tanθ wrong or use wrong geometric condition of hemisphere ?

help is appreciate

Thanks :!)
 

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Your tanθ should be equal to (3/5)r/(4/5)r=3/4

tanθ=v2/r'g
v2=3/4x3/5rg=3/4x3/5grx5/5=3.3.5gr/4.5.5
 
Oh! thanks azizlwl :wink:

i get it this is an easy one but i can't notice o:)
 

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