Help with a work/kinetic energy/potential energy problem

[haruspex]

ok so the radial acceleration is the vector gcos(theta) ?

If theta is still wrt the horizontal, no. At the top of the loop, the radial direction will be vertical.

 

[toesockshoe]

If theta is still wrt the horizontal, no. At the top of the loop, the radial direction will be vertical.

oh sorry. i meant gsin(theta). but how does this factor in with v^2/r?

 

[haruspex]

oh sorry. i meant gsin(theta). but how does this factor in with v^2/r?

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]

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?

it has to equal the weight of the mass?

 

[haruspex]

it has to equal the weight of the mass?

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?

 

[toesockshoe]

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?

it has to be greater than 0?

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.

 

[haruspex]

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?

 

[toesockshoe]

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.

 

[toesockshoe]

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.

 

[haruspex]

gsin(theta) = v^2/r

Yes!

and that is true for all moments on the sphere right? not just when the normal force is 0.

No.
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?

 

[toesockshoe]

Yes!

No.
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?

F_n- gsin(\theta) correct?

 

[haruspex]

F_n- gsin(\theta) 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?

 

[toesockshoe]

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?

 

[haruspex]

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?

 

[toesockshoe]

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
F_n - mgsin(\theta) = \frac{mv^2}{r} correct?

lets assume that the y-axis is pointing along the same direction as the normal force.

 

[haruspex]

oh. it is
F_n - mgsin(\theta) = \frac{mv^2}{r} 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.

 

[toesockshoe]

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?

 

[Raghav Gupta]

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}##

 

[toesockshoe]

ok test: $\frac{mv^2}{r}$ how come your code didnt convert into latex? its properly coded... i don't see any spaces.

 

[Raghav Gupta]

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.
When writing equivalent to ##