Mechanical problem -- a mass sliding down a curved ramp

AI Thread Summary
The discussion focuses on a physics problem involving an object sliding down a frictionless semicircular ramp. Key points include the application of energy conservation to determine the object's velocity at various points and the need to consider centripetal force to understand the conditions under which the object will leave the track. The relationship between gravitational force and normal force is emphasized, with the centripetal force requirement being crucial for the object's motion. The calculations lead to the conclusion that the height-to-radius ratio, h/R, is 5/2. The final steps involve using the object's velocity at point B to calculate its landing distance from point O after it leaves the ramp.
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Homework Statement


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An object which has mass = m is released from point A which has height of h. The object moves along the rail with no friction. The rail OB is a semicircular rail which C is the center and R is the radius. If the object falls from the rail at the point B, find the following.
1) h/R
2) Distance of the object from point O if the object lands on the surface without hitting the rail

Homework Equations

The Attempt at a Solution


Using the law of energy conservation

mgh + 1/2mv2 = mgh + 1/2mv2

The object should have velocity of sqrt(2gh) at point O

What should I do next at the semicircular rail?
How to calculate the reaction from the rail since the object will fall if the reaction is 0?
 

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What speed must the object have at B to just start leaving the track at that point?
 
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Doc Al said:
What speed must the object have at B to just start leaving the track at that point?
The vertical speed is 0?

I haven't studied the centripetal force yet. Is it used in the calculation?
 
PeppaPig said:
The vertical speed is 0?
What's the horizontal speed?

PeppaPig said:
I haven't studied the centripetal force yet. Is it used in the calculation?
Yes. To figure out the minimum speed it must have to just barely make it to point B, you must consider centripetal force & acceleration.
 
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FYI: If the object isn't moving fast enough, it will leave the track before getting to point B.
 
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Doc Al said:
FYI: If the object isn't moving fast enough, it will leave the track before getting to point B.

I just read about the centripetal force on wikipedia. The formula is F = mv2/R which v2/R is the centripetal acceleration.
How can I use this to solve this problem?
 
PeppaPig said:
The formula is F = mv2/R which v2/R is the centripetal acceleration.
Good. Your textbook should have plenty of examples to learn from as well.

PeppaPig said:
How can I use this to solve this problem?
Apply Newton's 2nd law at point B. What forces act?
 
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Doc Al said:
Good. Your textbook should have plenty of examples to learn from as well.Apply Newton's 2nd law at point B. What forces act?

There are the gravitational force and centripetal force which are mg and mv2/R. Reaction force = 0 means the centripetal force (from object to center) is equal to gravitational force. Is that correct?
 
PeppaPig said:
Reaction force = 0 means the centripetal force (from object to center) is equal to gravitational force. Is that correct?
Good! Now set up an equation to solve for the speed/energy needed at that point.
 
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Doc Al said:
Good! Now set up an equation to solve for the speed/energy needed at that point.

I have another question. v at point B is equal to v and point O?
 
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PeppaPig said:
I have another question. v at point B is equal to v and point O?
Why would you think that? Energy is conserved! (Those points are at different heights.)
 
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Doc Al said:
Why would you think that? Energy is conserved! (Those points are at different heights.)
Oh! I forgot about that.

1/2m(2gh) = 1/2mv2 + mg(2R)
v2 = 2g(h - 2R)

Then mg = mv2/R

g = (2g(h-2R))/R

h/R = 5/2

Then use v at point B to calculate distance from O after landing using projectile. Is that correct?
 
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PeppaPig said:
h/R = 5/2
Good!

PeppaPig said:
Then use v at point B to calculate distance from O after landing using projectile. Is that correct?
Exactly.
 
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Doc Al said:
Good!Exactly.
Thank you very much.
 
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PeppaPig said:
There are the gravitational force and centripetal force which are mg and mv2/R. Reaction force = 0 means the centripetal force (from object to center) is equal to gravitational force. Is that correct?
Yes but that's not quite the way I would think about it. The two forces aren't gravity and the centripetal force. The two forces are gravity and the normal force from the track.

In order to move in a circle something must provide a net centripetal force = mv^2/, no more and no less. In this case the two forces that combine to produce the centripetal force are gravity and the normal force produced by the track. The normal force originates from the inertia of the object which tries to make it move in a straight line.

If gravity provides less than mv^2/r then the normal force from the track will provide the rest.

If gravity provides more than mv^2/r the radius will reduce and it will fall down.
 

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