Block Sliding on Sphere

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Homework Statement



A large sphere was built, with a radius of about 5. Imagine that such a sphere has a radius "R" and a frictionless surface. A small block of mass "M" slides starts from rest at the very top of the sphere and slides along the surface of the sphere. The block leaves the surface of the sphere when it reaches a height "H(critical)" above the ground.

Using Newton's 2nd law, find v(critical), the speed of the block at the critical moment when the block leaves the surface of the sphere.

Assume that the height at which the block leaves the surface of the sphere is H(critical).

Express the speed in terms of R, H(critical), and g (the magnitude of the accleration due to gravity). Do not use (theta) in your answer.


Homework Equations



None given.

The Attempt at a Solution



I know that the moment that the block leaves the surface of the sphere the normal force will equal 0, so when this happens the gravitational force (it's radial component) will be the only force causing centripetal acceleration.

To find the radial component I decomposed the vector and found that the radial component of the gravitational force was "M*g*sin(theta)". Since the centripetal acceleration of the block at a speed v is (v^2)/R, I thought I could simply set them equal to eachother and then solve for v. This gave me an answer of sqrt(M*g*R*sin(theta))=v. But I cannot use theta or M in my answer, I tried to set up a right triangle to relate sin(theta) to R and H(critical), I found that sin(theta) was equal to h(of block)/R however the computer program I use to submit my answers is telling me I'm off by an additive constant. I cannot think of a way to express sin(theta) in only terms of h and R. Also if I did manage to get past that I am unsure of how I would be able to get rid of M (express it in other terms).

I think I might be on the wrong track with this one.

Thanks.
 

Answers and Replies

  • #2
PSOA
You shouldn't head that way to solve the problem because of that angle and the mass of the block. I will give you a big HINT: Conservation of Mechanical Energy.
 
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  • #3
Doc Al
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Using Newton's 2nd law, find v(critical), the speed of the block at the critical moment when the block leaves the surface of the sphere.

Assume that the height at which the block leaves the surface of the sphere is H(critical).

Express the speed in terms of R, H(critical), and g (the magnitude of the accleration due to gravity). Do not use (theta) in your answer.
I find this problem statement puzzling. On the one hand, you are asked to use Newton's 2nd law to find v(critical), but on the other hand you are told to assume the critical height is H(critical). If you already know H(critical), then finding the v(critical) is simple using PSOA's big hint: conservation of energy. No need for any analysis using Newton's 2nd law.

I suspect someone botched this problem statement.

Generally, one is asked to find the critical angle (or critical height--given one you can find the other) and the critical speed. That problem requires using Newton's 2nd law--like you did--as well as conservation of energy.

I know that the moment that the block leaves the surface of the sphere the normal force will equal 0, so when this happens the gravitational force (it's radial component) will be the only force causing centripetal acceleration.
Good.

To find the radial component I decomposed the vector and found that the radial component of the gravitational force was "M*g*sin(theta)". Since the centripetal acceleration of the block at a speed v is (v^2)/R, I thought I could simply set them equal to eachother and then solve for v. This gave me an answer of sqrt(M*g*R*sin(theta))=v.
Right approach, but you made an error: That M shouldn't be there.

Combine this with conservation of energy and you can solve for theta and specify v purely in terms of R and g--no need to assume a value for H(critical).
 
  • #4
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Thanks for your help.

Right approach, but you made an error: That M shouldn't be there.

Combine this with conservation of energy and you can solve for theta and specify v purely in terms of R and g--no need to assume a value for H(critical).
How do you rid yourself of the M??? Surely simply eliminating it from the equation would not be possible. For conservation of energy,

K_i + U_i = K_f + U_f

0 (block starts from rest) + mg2R (h=2R)= (1/2)mv(crit)^2 + mgh(crit)

v = sqrt(2(2gR - gh(crit)))

This is the answer for the second portion of the problem which I have already correctly solved.

The part of the question I am still stuck on is asking for me to use Newton's second law to find the speed (v(crit)) of the block at the critical moment when the block leaves the surface of the globe. Assuming that the height at which the block leaves the surface of the globe is h(crit). I am only allowed to give my answer in terms of R, h(crit) and g.

Following the hints that I am given I have found the magnitude of the radial component of the gravitational force on the block to be v^2/r and the magnitude of the radial component of the gravitational force (Fr) on the block to be mgsin(theta).

The last hint that they are giving me is

"Having found (Fr)=mgsin(theta) , you now need to find sin(theta) in terms of h (the height of the block) and R. You need to find a right triangle where that is the included angle and R is the hypoteneuse. Using this triangle, what is sin theta? In terms of h and R."

I used sin= opp/hyp. Set my hyp to the radius R then the opposite side would be mgsin(theta). sin theta would be h/R. But they are telling me that I am off my an additive constant. I have tried all the ways I could think of to add or subract h or R, but no luck.

Thanks.
 
  • #5
PSOA
Canadian, note that [tex]\Sigma F_c=mg\sin\theta=mv^2/r[/tex].
 
  • #6
Doc Al
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How do you rid yourself of the M??? Surely simply eliminating it from the equation would not be possible.
The reason the M did not simply drop out of your equations--like it should--is that you have mistaken the centripetal acceleration, v^2/R, for the centripetal force.
For conservation of energy,

K_i + U_i = K_f + U_f

0 (block starts from rest) + mg2R (h=2R)= (1/2)mv(crit)^2 + mgh(crit)

v = sqrt(2(2gR - gh(crit)))

This is the answer for the second portion of the problem which I have already correctly solved.
OK.

The part of the question I am still stuck on is asking for me to use Newton's second law to find the speed (v(crit)) of the block at the critical moment when the block leaves the surface of the globe. Assuming that the height at which the block leaves the surface of the globe is h(crit). I am only allowed to give my answer in terms of R, h(crit) and g.
Again, I find this problem poorly posed, since using Newton's 2nd law (and conservation of energy) one can solve for v without having to know h(crit) ahead of time.

Following the hints that I am given I have found the magnitude of the radial component of the gravitational force on the block to be v^2/r and the magnitude of the radial component of the gravitational force (Fr) on the block to be mgsin(theta).
Note that v^2/r is an acceleration, not a force.

The last hint that they are giving me is

"Having found (Fr)=mgsin(theta) , you now need to find sin(theta) in terms of h (the height of the block) and R. You need to find a right triangle where that is the included angle and R is the hypoteneuse. Using this triangle, what is sin theta? In terms of h and R."
OK, for whatever reason they want you to do a little trig to change variables to eliminate sin(theta). (Again, nothing stops you from solving for the actual value of theta--but you may as well follow the instructions.)

I used sin= opp/hyp. Set my hyp to the radius R then the opposite side would be mgsin(theta). sin theta would be h/R. But they are telling me that I am off my an additive constant. I have tried all the ways I could think of to add or subract h or R, but no luck.
Here's a hint: Realize that this is a sphere, not a hemisphere. And that theta = 0 would represent a horizontal line going through the center of the sphere--not at ground level. Is the critical point above or below that line?
 
  • #7
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Thanks, that last little hint did it. I really appreciate your help.
 
  • #8
PSOA
Docal Al very good explanation, I think I am not prepared right now to help people.
 

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