Block sliding down frictionless semicircle

  • #1

Homework Statement


A block of mass, m, sits atop a semicircular bowl of radius, r, and the angle the radius makes with the horizontal is [tex]\theta[/tex]. Find what angle, [tex]\theta[/tex], the block will slide off the bowl.


Homework Equations


[tex]\frac{1}{2}mv^2 + mgy = \frac{1}{2}mv^2 + mgy[/tex]

[tex]\alpha=ar[/tex]

[tex]a_{c}=\frac{v^2}{r}[/tex]

The Attempt at a Solution


Drawing a free-body diagram, the forces I think at work are the normal force, gravity, and possibly a centripetal force. The resolution of gravity along a circular path is where my trouble begins. I don't think it's really equivalent to resolve gravity as the sum of two x and y vectors along an incline, since this is a circle. Maybe resolving into x=rcos[tex]\theta[/tex] and y=rsin[tex]\theta[/tex] and saying mgrcos[tex]\theta[/tex]=X motion and Y motion is mgrsin[tex]\theta[/tex]=y. Then I'm stuck how to incorporate the normal force into either of those two equations (assuming they're accurate representations of the motion of this object along this semicircle). Maybe [tex]\sum{F_{y}}=F_{n}-mgrsin\theta=mar[/tex] and [tex]\sum{F_{x}}=mgrcos\theta=mar[/tex]. I guess if I could by some mathematical bastardization equate mgrcos[tex]\theta[/tex]=mgrsin[tex]\theta[/tex] I'd get that [tex]\theta=\frac{\Pi}{4}[/tex]

I hope you all can give me a little guidance so that I can solve this problem. I know it's been posted before, but I searched, and I couldn't any help toward a solution.
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
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What's the normal force acting on the block as it slides off the bowl?
 
  • #3
so if the normal force equals zero then [tex]-mgrsin\theta=mgrcos\theta[/tex] and [tex]\theta=\frac{-\pi}{4}[/tex] which would have to become positive [tex]\frac{\pi}{4}[/tex] since...?
 
  • #4
Doc Al
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so if the normal force equals zero then [tex]-mgrsin\theta=mgrcos\theta[/tex] and [tex]\theta=\frac{-\pi}{4}[/tex] which would have to become positive [tex]\frac{\pi}{4}[/tex] since...?
The normal force is zero as it loses contact with the surface, but why in the world are you attempting to equate "vertical" and "horizontal" components of gravity? (At least I think that's what you're doing. To properly find the components of gravity, draw yourself a diagram and apply a bit of trig. Don't convert coordinates using x = r cosθ.)

Hint: Apply Newton's 2nd law in the radial direction.
 
  • #5
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The most important equation is

[tex] N = 0 [/tex]

At theta what are the forces on the block in the radial direction? Equate them as N is zero.

You will get a relation in r g and theta and velocity. Find the velocity by energy conservation
 
  • #6
ac=gcos(theta)
 

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