Maximizing Speed on a Snow-Covered Hill: Solving for Angle and Maximum Speed

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The discussion revolves around a physics problem involving a sled on a frictionless, hemispherical snow-covered hill. The key points include finding the sled's speed at a given angle "theta" using energy conservation, where potential energy converts to kinetic energy. Newton's laws are applied to determine the maximum speed without the sled leaving the surface, with centripetal acceleration being a critical factor. The height of the sled is expressed as a function of theta, leading to the equation V = sqrt(g R sin theta). The conversation emphasizes the importance of analyzing forces and the relationship between height and angle in this context.
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the problem states as this
it's a energy related problem

A sled starts from rest at the top of the frictionless, hemispherical snow-covered hill.
a)Find an expression for the sled's speed when it is at angle "theta"
b)Use Newton's laws to find the maximum speed the sled can have at angle "theta" without leaving the surface.
c)at what angle "theta"max does the sled "fly off" the hill?

since the surface is fristionless the energy is conserved we can say E at start = E at angle theta
now i realize at the top the sled has zero kinetic energy and Pe=mgR
and the at point "thete" Ke=1/2mV^2 and Pe=mgh

now at this point I am stuck and not sure what to do with all this data

thanks in advance
 

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Since only gravity is doing work, the potential energy of the sled is converted to kinetic energy.
Try to find the height of the sled as a function of theta.
 
Also remember that only gravity can provide the necessary centripetal acceleration..
 
Don't forget to use Newton's 2nd law to analyze the forces on the sled. Since it follows a circular path (until it leaves the surface), apply what you know about centripetal acceleration.
 
would the height be equal to h=r sin(theta)?
 
well considering centripedal forces i have
a=V^2/rsin theta
Fn-mg=ma
Fn-mg=mV^2/R sin theta
Fn=mg-mV^2/R sin theta
0=mg-mV^2/R sin theta
mV^2=mgRsin theta
V= sqrt(g Rsin theta)
 
vbtweakin said:
would the height be equal to h=r sin(theta)?
The angle is measured with respect to the vertical (in your diagram), so your height should have maximum value when the angle is zero. Does the sine function give you that?
 
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