Help with a work/kinetic energy/potential energy problem

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In summary, the problem involves a point mass sliding down a frictionless solid sphere and finding the angle at which it flies off the surface. The necessary equations to solve the problem are the work-energy theorem, the gravitational potential energy equation, and the kinetic energy equation. The solution involves finding the acceleration of the block, using the fact that at the point of leaving the surface, there is no friction and normal force, so the only force acting is the weight, mg. The acceleration can be calculated by equating the radial component of g to the centripetal acceleration. The angle can then be found using the centripetal acceleration equation and conservation of mechanical energy.
  • #36
haruspex said:
You've established that the radial acceleration is g sin(theta). What does the radial acceleration need to be to stay in contact with the curved surface? What relationship do you think that leads to?
toesockshoe said:
it has to be greater than 0?
haruspex said:
A weight is a force, not an acceleration.
Let me ask you something - what do you think centripetal acceleration means? How would you define it?

its is the acceleration component that is point towards the center of the circular motion it is traveling in. that's all i know.
 
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  • #37
toesockshoe said:
it is the acceleration component that is point towards the center of the circular motion it is traveling in. that's all i know.
Right, it is the radial component of the acceleration. We know the centripetal acceleration is ##\frac {v^2}r##, and in post #33 I wrote that you had shown the radial component of acceleration to be ##g\sin(\theta)##. So what equation does that give you?
 
  • #38
haruspex said:
Right, it is the radial component of the acceleration. We know the centripetal acceleration is ##\frac {v^2}r##, and in post #33 I wrote that you had shown the radial component of acceleration to be ##g\sin(\theta)##. So what equation does that give you?
it means gsin(theta) = v^2/r.
 
  • #39
haruspex said:
Right, it is the radial component of the acceleration. We know the centripetal acceleration is ##\frac {v^2}r##, and in post #33 I wrote that you had shown the radial component of acceleration to be ##g\sin(\theta)##. So what equation does that give you?
and that is true for all moments on the sphere right? not just when the normal force is 0.
 
  • #40
toesockshoe said:
gsin(theta) = v^2/r
Yes! :smile:
toesockshoe said:
and that is true for all moments on the sphere right? not just when the normal force is 0.
No. :frown:
To maintain contact, the radial component of the net force must be ##\frac{mv^2}r##.
In terms of gravity and the normal force (the only two forces present), what is the radial component of the net force?
 
  • #41
haruspex said:
Yes! :smile:

No. :frown:
To maintain contact, the radial component of the net force must be ##\frac{mv^2}r##.
In terms of gravity and the normal force (the only two forces present), what is the radial component of the net force?
[tex] F_n- gsin(\theta) [/tex] correct?
 
  • #42
toesockshoe said:
[tex] F_n- gsin(\theta) [/tex] correct?
Yes, taking radially outward as positive and g as positive. So what equation can you write for staying in contact with the sphere when the normal might not be zero?
 
  • #43
haruspex said:
Yes, taking radially outward as positive and g as positive. So what equation can you write for staying in contact with the sphere when the normal might not be zero?
im not 100 percent sure what your asking. can you elaborate?
 
  • #44
toesockshoe said:
im not 100 percent sure what your asking. can you elaborate?
In post #38 you wrote the correct equation relating gravitational acceleration to centripetal acceleration for the case where contact is just maintained, i.e. where the normal force is zero.
What is the equation for the case where the normal force is not zero?
 
  • #45
haruspex said:
In post #38 you wrote the correct equation relating gravitational acceleration to centripetal acceleration for the case where contact is just maintained, i.e. where the normal force is zero.
What is the equation for the case where the normal force is not zero?
oh. it is
[tex] F_n - mgsin(\theta) = \frac{mv^2}{r} [/tex] correct?

lets assume that the y-axis is pointing along the same direction as the normal force.
 
  • #46
toesockshoe said:
oh. it is
[tex] F_n - mgsin(\theta) = \frac{mv^2}{r} [/tex] correct?

lets assume that the y-axis is pointing along the same direction as the normal force.
Close, but you have a sign error. The normal force points radially outward, but the centripetal acceleration is radially inward. And remember, when the normal force is zero you should get the same equation as in post #38.
 
  • #47
haruspex said:
Close, but you have a sign error. The normal force points radially outward, but the centripetal acceleration is radially inward. And remember, when the normal force is zero you should get the same equation as in post #38.
oh that's right. so I should switch the Fn and the mgsin(theta) around and make my y-axis point upward in the direction as the gravity force. also, whenever I use latex, the latex forumla gets placed in the center of the page. can you give me a sample code of how to put the latex code in line with your normal text? for example, here is how i would type a regular latex code [ tex ] F_n-mgsin(\theta)=\frac{mv^2}{r} [ / tex]. why does that become centered?
 
  • #48
toesockshoe said:
oh that's right. so I should switch the Fn and the mgsin(theta) around and make my y-axis point upward in the direction as the gravity force. also, whenever I use latex, the latex forumla gets placed in the center of the page. can you give me a sample code of how to put the latex code in line with your normal text? for example, here is how i would type a regular latex code [ tex ] F_n-mgsin(\theta)=\frac{mv^2}{r} [ / tex]. why does that become centered?
Use itex instead of tex.
Also tex is equivalent to $$ and itex is equivalent to ##
Type ##\frac{mv^2}{r}##
 
  • #49
ok test: ##\frac{mv^2}{r} ## how come your code didnt convert into latex? its properly coded... i don't see any spaces.
 
  • #50
toesockshoe said:
ok test: ##\frac{mv^2}{r} ## how come your code didnt convert into latex? its properly coded... i don't see any spaces.
Since I have done the trickery of using ## before that whole code.:wink:
When writing equivalent to ##
 
<h2>1. How do I calculate work in a work/kinetic energy/potential energy problem?</h2><p>In order to calculate work, you need to multiply the force applied to an object by the distance the object moves in the direction of the force. The formula for work is W = Fd, where W represents work, F represents force, and d represents distance.</p><h2>2. What is the difference between kinetic energy and potential energy?</h2><p>Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state. Kinetic energy is dependent on an object's mass and velocity, while potential energy is dependent on an object's height or distance from a reference point.</p><h2>3. How do I calculate kinetic energy in a work/kinetic energy/potential energy problem?</h2><p>Kinetic energy can be calculated using the formula KE = 1/2mv^2, where KE represents kinetic energy, m represents mass, and v represents velocity. Make sure to use consistent units when plugging in values for this formula.</p><h2>4. Can potential energy be negative in a work/kinetic energy/potential energy problem?</h2><p>Yes, potential energy can be negative in certain situations. For example, if an object is below a reference point and has gravitational potential energy, it will have a negative value since it has the potential to fall and release energy. However, in most cases, potential energy is positive.</p><h2>5. How do I use the work-energy theorem in a work/kinetic energy/potential energy problem?</h2><p>The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This can be written as Wnet = ΔKE. This theorem can be used to solve for unknown values in a problem involving work, kinetic energy, and potential energy.</p>

1. How do I calculate work in a work/kinetic energy/potential energy problem?

In order to calculate work, you need to multiply the force applied to an object by the distance the object moves in the direction of the force. The formula for work is W = Fd, where W represents work, F represents force, and d represents distance.

2. What is the difference between kinetic energy and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state. Kinetic energy is dependent on an object's mass and velocity, while potential energy is dependent on an object's height or distance from a reference point.

3. How do I calculate kinetic energy in a work/kinetic energy/potential energy problem?

Kinetic energy can be calculated using the formula KE = 1/2mv^2, where KE represents kinetic energy, m represents mass, and v represents velocity. Make sure to use consistent units when plugging in values for this formula.

4. Can potential energy be negative in a work/kinetic energy/potential energy problem?

Yes, potential energy can be negative in certain situations. For example, if an object is below a reference point and has gravitational potential energy, it will have a negative value since it has the potential to fall and release energy. However, in most cases, potential energy is positive.

5. How do I use the work-energy theorem in a work/kinetic energy/potential energy problem?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This can be written as Wnet = ΔKE. This theorem can be used to solve for unknown values in a problem involving work, kinetic energy, and potential energy.

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