Help with a work/kinetic energy/potential energy problem

[toesockshoe]

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

[/B]

 

[insightful]

I don't think that 'r' can be in the answer since the object of the inverse sine must be from 0 to 1.

 

[toesockshoe]

So do u know where I am going wrong?

 

[toesockshoe]

Oh I think I figured it out

 

[haruspex]

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.

 

[toesockshoe]

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?

 

[haruspex]

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?

 

[insightful]

I find it helpful to look at limits. What centripetal acceleration is needed at theta=90?...at theta=0?

 

[eifphysics]

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.

 

[toesockshoe]

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)
?

 

[toesockshoe]

ok, I am not really sure how to find the acceleration. can anyone give me a hint on how to solve this problem?

 

[insightful]

When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface (g sin(theta)).

 

[toesockshoe]

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?

 

[eifphysics]

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.

 

[toesockshoe]

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?

 

[eifphysics]

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.

 

[SammyS]

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 !

 

[haruspex]

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..

 

[insightful]

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.

 

[toesockshoe]

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) ?

 

[toesockshoe]

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) ?

 

[haruspex]

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.

 

[toesockshoe]

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?

 

[haruspex]

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 (F_g).

 

[toesockshoe]

Right, assuming Fg means force due to gravity (F_g).

yes, so centripital acceleration would just be the sum Fgcos(theta) and Fg(sing theta)? if so, where does the v^2/R factor in?

 

[haruspex]

yes, so centripital 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:

the radial acceleration is Fgsin(theta)

bearing in mind that that should read: the radial acceleration is Fgsin(theta)/m

 

[toesockshoe]

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 ??

 

[toesockshoe]

but then it owuldnt even point to the center.

 

[haruspex]

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.

 

[toesockshoe]

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) ?

 

[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 ##