- #1
XJellieBX
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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 an angle φ.
Usind the conservation of energy, I found this expression to be v=[tex]\sqrt{2g(R-Rcosφ}[/tex].
b) Use Newton's laws to find the max speed the sled can have at angle φ without leaving the surface.
I used the fact that F_net=ma --> [tex]\sum[/tex]F_r=n_r+F_g,r=(mv^2)/R=0
So then, F_r=n=mg/cosφ --> (mv^2)/R = mg/cosφ => v=[tex]\sqrt{\frac{Rg}{cosφ}}[/tex].
This one I'm not sure the cosφ in the denominator is right. But the general format corresponds to the max orbital speed. I'm not certain the derivation is formal enough.
c) At what angle φ_max does the sled "fly off" the hill?
I'm not sure how to approach this one. I was thinking that maybe it's when the expression in part b is undefined, but then that gives me π/2. Which is when the sled is at the bottom of the hill.
Any ideas/suggestions would be appreciated.
a) Find an expression for the sled's speed when it is at an angle φ.
Usind the conservation of energy, I found this expression to be v=[tex]\sqrt{2g(R-Rcosφ}[/tex].
b) Use Newton's laws to find the max speed the sled can have at angle φ without leaving the surface.
I used the fact that F_net=ma --> [tex]\sum[/tex]F_r=n_r+F_g,r=(mv^2)/R=0
So then, F_r=n=mg/cosφ --> (mv^2)/R = mg/cosφ => v=[tex]\sqrt{\frac{Rg}{cosφ}}[/tex].
This one I'm not sure the cosφ in the denominator is right. But the general format corresponds to the max orbital speed. I'm not certain the derivation is formal enough.
c) At what angle φ_max does the sled "fly off" the hill?
I'm not sure how to approach this one. I was thinking that maybe it's when the expression in part b is undefined, but then that gives me π/2. Which is when the sled is at the bottom of the hill.
Any ideas/suggestions would be appreciated.