Help with a work/kinetic energy/potential energy problem

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SUMMARY

The discussion centers on solving a physics problem involving a point mass sliding down a frictionless solid sphere and determining the angle at which the mass flies off. Key equations include gravitational potential energy (mgh) and kinetic energy (1/2 mvf^2 - 1/2 mvi^2). Participants emphasize the importance of centripetal acceleration, noting that at the point of leaving the surface, the centripetal acceleration (v^2/r) must equal the gravitational component normal to the surface (g sin(θ)). The conversation highlights the need for clarity in problem-solving and the use of free body diagrams to visualize forces.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy equations
  • Familiarity with centripetal acceleration concepts
  • Ability to analyze forces using free body diagrams
  • Knowledge of trigonometric functions in physics contexts
NEXT STEPS
  • Study the relationship between centripetal acceleration and gravitational forces
  • Learn how to apply conservation of mechanical energy in dynamic systems
  • Explore the use of free body diagrams in solving physics problems
  • Investigate the implications of normal force in circular motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for insights into teaching concepts of energy and motion in circular paths.

toesockshoe
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Homework Statement



A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r. Find the angle at which the mass flies off the sphere.

Homework Equations



w=delta energy
Gravitaional potential energy = mgh
Kinetic energy = 1/2mvf^2 - 1/2mvi^2[/B]

The Attempt at a Solution



is my answer correct?

Please check my picture
physics.jpg
[/B]
 
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I don't think that 'r' can be in the answer since the object of the inverse sine must be from 0 to 1.
 
So do u know where I am going wrong?
 
Oh I think I figured it out
 
Please do not post working as an image. It makes it hard to draw your attention to a particular line it.
Your centripetal acceleration calculation doesn't seem to take into account the directions of the forces. It is not in the same direction as g.
On the left hand side, the expression ##(r-\sin(\theta))## should raise alarm bells. It makes no sense dimensionally.
 
haruspex said:
Please do not post working as an image. It makes it hard to draw your attention to a particular line it.
Your centripetal acceleration calculation doesn't seem to take into account the directions of the forces. It is not in the same direction as g.
On the left hand side, the expression ##(r-\sin(\theta))## should raise alarm bells. It makes no sense dimensionally.
yes... the correct equation shoudl be

r-rsin(\theta)

correct? and the rest of the problem should be correct?
 
toesockshoe said:
yes... the correct equation shoudl be

r-rsin(\theta)

correct? and the rest of the problem should be correct?
No, there's still the problem with your centripetal acceleration equation, as mentioned. Which direction is that in? How does that compare with the direction of g?
 
I find it helpful to look at limits. What centripetal acceleration is needed at theta=90?...at theta=0?
 
You should be able to solve the problem with two unknowns: v and θ, and two equations: one from the centripetal motion constraint and one from conservation of mechanical energy. It looks like you need to make you problem solving process more clear and organized.
 
  • #10
haruspex said:
No, there's still the problem with your centripetal acceleration equation, as mentioned. Which direction is that in? How does that compare with the direction of g?
so is
a_c=\frac{mv^2}{r}+gcos(\theta)
?
 
  • #11
ok, I am not really sure how to find the acceleration. can anyone give me a hint on how to solve this problem?
 
  • #12
When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface (g sin(theta)).
 
  • #13
insightful said:
When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface (g sin(theta)).
why?
 
  • #14
Because the ball is moving around the sphere, velocity is tangent to the sphere, so centripetal force must be normal to the sphere surface. When the ball flies off, it means velocity is too big and even the largest possible centripetal force cannot maintain the circular motion. The largest centripetal force possible is component of gravity normal to the surface(with normal force equals 0). This is why When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface.
 
  • #15
eifphysics said:
Because the ball is moving around the sphere, velocity is tangent to the sphere, so centripetal force must be normal to the sphere surface. When the ball flies off, it means velocity is too big and even the largest possible centripetal force cannot maintain the circular motion. The largest centripetal force possible is component of gravity normal to the surface(with normal force equals 0). This is why When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface.
ok by what is the acceleration of the block. In other words... for:
F_T-F_g=ma

what is a?
 
  • #16
a = v2/r = gcosθ. gcosθ is the largest centripetal force, component of gravity normal to the sphere surface. If θ is larger v will be too big to maintain in circular motion. Then use conservation of mechanical energy to obtain a second equation for v and θ.
 
  • #17
toesockshoe said:

Homework Statement



A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r. Find the angle at which the mass flies off the sphere.

Homework Equations



w=delta energy
Gravitaional potential energy = mgh
Kinetic energy = 1/2mvf^2 - 1/2mvi^2[/B]

The Attempt at a Solution



is my answer correct?

Please check my picture https://www.physicsforums.com/attachments/physics-jpg.83522/
[ ATTACH=full]83522[/ATTACH] [/B]
As is almost always the case, a Free Body Diagram is very useful, if not absolutely essential !
 
  • #18
toesockshoe said:
ok by what is the acceleration of the block. In other words... for:
F_T-F_g=ma

what is a?
There's no friction, and at the point of leaving the surface there's no normal force, so the only force is mg. But that does not mean g equals the centripetal acceleration. The mass can have both tangential and radial acceleration. We don't care what the tangential acceleration is. We need to extract the radial component of g and equate that to the centripetal acceleration..
 
  • #19
eifphysics said:
a = v2/r = gcosθ. gcosθ is the largest centripetal force, component of gravity normal to the sphere surface. If θ is larger v will be too big to maintain in circular motion.
This looks like you're measuring theta from the vertical. I believe OP is measuring it from the horizontal.
 
  • #20
haruspex said:
No, there's still the problem with your centripetal acceleration equation, as mentioned. Which direction is that in? How does that compare with the direction of g?
ok so measuring theta from the horizonta... isn't a_c = \frac{v^2}{r} + gsin(\theta) or does the gravitational force cause the centripial acceleraion in which
a_c = \frac{v^2}{r} = gsin(\theta) ?
 
  • #21
SammyS said:
As is almost always the case, a Free Body Diagram is very useful, if not absolutely essential !
ok so measuring theta from the horizonta... isn't a_c = \frac{v^2}{r} + gsin(\theta) or does the gravitational force cause the centripial acceleraion in which
a_c = \frac{v^2}{r} = gsin(\theta) ?
 
  • #22
toesockshoe said:
ok so measuring theta from the horizonta... isn't a_c = \frac{v^2}{r} + gsin(\theta) or does the gravitational force cause the centripial acceleraion in which
a_c = \frac{v^2}{r} = gsin(\theta) ?
Centripetal force is a resultant force, not an applied force. It is the radial component of the sum of all applied forces.
 
  • #23
haruspex said:
Centripetal force is a resultant force, not an applied force. It is the radial component of the sum of all applied forces.
alright, so i split up gravity into tangental and radial components. at any point on the sphere, theta above the horizontal, the radial acceleration is Fgsin(theta) and the tangential acceleration is Fgcos(theta) correct?
 
  • #24
toesockshoe said:
alright, so i split up gravity into tangental and radial components. at any point on the sphere, theta above the horizontal, the radial acceleration is Fgsin(theta) and the tangential acceleration is Fgcos(theta) correct?
Right, assuming Fg means force due to gravity (Fg).
 
  • #25
haruspex said:
Right, assuming Fg means force due to gravity (Fg).
yes, so centripetal acceleration would just be the sum Fgcos(theta) and Fg(sing theta)? if so, where does the v^2/R factor in?
 
  • #26
toesockshoe said:
yes, so centripetal acceleration would just be the sum Fgcos(theta) and Fg(sing theta)? if so, where does the v^2/R factor in?
First, a small correction: those are forces, not accelerations.
Now a bigger correction: the (vector) sum of Fgcos(theta) and Fg(sin theta) is the total applied force.
The centripetal acceleration is the radial component of the total acceleration.
Combine those facts with what you posted earlier:
toesockshoe said:
the radial acceleration is Fgsin(theta)
bearing in mind that that should read: the radial acceleration is Fgsin(theta)/m
 
  • #27
haruspex said:
First, a small correction: those are forces, not accelerations.
Now a bigger correction: the (vector) sum of Fgcos(theta) and Fg(sin theta) is the total applied force.
The centripetal acceleration is the radial component of the total acceleration.
Combine those facts with what you posted earlier:

bearing in mind that that should read: the radial acceleration is Fgsin(theta)/m
alright so the radial acceleration would be...: the sum of vector Fgcos(tehta)/m + vector Fgsin(tehta)/m ??
 
  • #28
but then it owuldnt even point to the center.
 
  • #29
toesockshoe said:
alright so the radial acceleration would be...: the sum of vector Fgcos(tehta)/m + vector Fgsin(tehta)/m ??
No, the radial component of acceleration results from the radial component of force. Only one of those is the radial component.
 
  • #30
haruspex said:
No, the radial component of acceleration results from the radial component of force. Only one of those is the radial component.
ok so the radial acceleration is the vector gcos(theta) ?
 

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