| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
10 Dec 2007 01:05:20 PM |
| Object: |
Angular and linear velocity |
Suppose I have a ball resting on a horizontal platform. Now I apply an
impulse P such that the ball starts rolling with a velocity u without
slipping. The moment of inertia of the ball is 2/5*m*a^2 where a is
the radius of the ball. Now how do I calculate the impulse P and how
do I find out where that impulse was applied?
Concerning the second part of the question: I'd say that it has to be
applied above the center of mass of the ball. Is that right?
Concerning the first part: I am confused. I think that the linear
velocity is given by v = P * m. So is P = u/m just the answer to the
question? Then again, we still have rotational motion. Here, I would
say that a * P = 2/5*m*a^2 * w, where w is the angular velocity. If I
then use w = v * a and use both equations above, I get 2/5 = 1.
Something's obviously wrong.
Can somebody please help out?
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| User: "Randy Poe" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 03:18:19 PM |
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On Dec 10, 2:05 pm, wrote:
Suppose I have a ball resting on a horizontal platform. Now I apply an
impulse P such that the ball starts rolling with a velocity u without
slipping. The moment of inertia of the ball is 2/5*m*a^2 where a is
the radius of the ball. Now how do I calculate the impulse P and how
do I find out where that impulse was applied?
Concerning the second part of the question: I'd say that it has to be
applied above the center of mass of the ball. Is that right?
Yes. If you apply it at the level of the center of the mass,
there is no torque and so the ball will start out by sliding.
Since you are told there is no slipping, then there must
be a torque applied which gives it rotation in the correct
direction.
Concerning the first part: I am confused. I think that the linear
velocity is given by v = P * m. So is P = u/m just the answer to the
question?
Yes. Conservation of (linear) momentum.
Then again, we still have rotational motion. Here, I would
say that a * P = 2/5*m*a^2 * w, where w is the angular velocity. If I
then use w = v * a and use both equations above, I get 2/5 = 1.
Something's obviously wrong.
Why did you use a*P on the left-side of this
equation?
Can somebody please help out?
Draw a picture. The impulse acts at a certain point
on the sphere (unknown distance above the CM).
The force is not perpendicular to the sphere's surface
at that point. What is the correct expression for
angular momentum?
- Randy
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| User: "Timo Nieminen" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 04:30:12 PM |
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On Mon, 10 Dec 2007, Randy Poe wrote:
On Dec 10, 2:05 pm, wrote:
Suppose I have a ball resting on a horizontal platform. Now I apply an
impulse P such that the ball starts rolling with a velocity u without
slipping. The moment of inertia of the ball is 2/5*m*a^2 where a is
the radius of the ball. Now how do I calculate the impulse P and how
do I find out where that impulse was applied?
Concerning the second part of the question: I'd say that it has to be
applied above the center of mass of the ball. Is that right?
Yes. If you apply it at the level of the center of the mass,
there is no torque and so the ball will start out by sliding.
Since you are told there is no slipping, then there must
be a torque applied which gives it rotation in the correct
direction.
Friction can allow some variation in the height at which the impulse is
applied. No friction, expect only one height possible.
In this case, it isn't sufficient to know the impulse - you need to know
the actual force acting, to make sure it's less than the maximum static
friction. So, this problem looks implicitly frictionless, even if not
explicitly so.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 03:49:07 PM |
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Thanks for your answer.
It is said that the impulse is horizontal. You are right, I used a * P
because I assumed that the impulse was applied at the top. However,
we're not told so. So let's say that it is applied at a distance b
from the center of mass. Then b * P = 2/5*m*a^2 * w (that's because b
and P are perpendicular to each other). Is that correct? Combining
with P = m * u from my first equation and using w = v * a I get b =
2/5 * a^3 - is that sensible? And what is the answer to the question
(how big is the impulse)?
Thanks for your help.
.
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| User: "Androcles" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 04:24:13 PM |
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<Whatever5k@web.de> wrote in message
news:d4f5452b-d6ae-4885-a060-49675dc5a04c@e25g2000prg.googlegroups.com...
: Thanks for your answer.
:
: It is said that the impulse is horizontal.
Ok, parallel to the horizon, 90 degrees from vertical.
:You are right, I used a * P
: because I assumed that the impulse was applied at the top. However,
: we're not told so.
This is a ball resting on a pool or billiard table, right?
If you apply a horizontal impulse to the top your cue
will miss the ball.
If you apply a horizontal impulse to the exact centre
the ball will slide forward. Friction between the table and
the ball will eventually dominate and the ball will roll.
If you apply a horizontal impulse to a point halfway between
the table and the centre of the ball, the ball will slide forward
and spin backwards. In the USA that is called putting "english"
on the ball, in England it is called putting spin in the ball, top on
the ball, side on the ball...
You can apply it to left, right, top, bottom, any point off-centre,
but I do not recommend you go to the extreme.
You can also tilt the cue so that the impulse is not horizontal.
The skillful pool, snooker or billiards player also knows to
"follow through", he doesn't tap the ball and withdraw the cue,
he allows it to continue forward with the tip of the cue remaining
in contact with the ball.
Successful golf players do much the same thing, allow the club
to complete the swing. The terminology is different, golfers
hook or slice or hit the "sweet" spot, but the physics is the same.
Good luck with your mathematics but remember it is only
a description of reality, not reality itself.
: So let's say that it is applied at a distance b
: from the center of mass. Then b * P = 2/5*m*a^2 * w (that's because b
: and P are perpendicular to each other). Is that correct? Combining
: with P = m * u from my first equation and using w = v * a I get b =
: 2/5 * a^3 - is that sensible? And what is the answer to the question
: (how big is the impulse)?
:
: Thanks for your help.
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| User: "Dirk Van de moortel" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 04:35:14 PM |
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"Androcles" <Engineer@hogwarts.physics_a> wrote in message news:h0j7j.26873$kt3.19439@fe3.news.blueyonder.co.uk...
[snip]
Good luck with your mathematics but remember it is only
a description of reality, not reality itself.
remember it is only Boolean algebra:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XOROnceMore.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORrevisited.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORContinued.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORpersistence.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORWildStab.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LooksBoolean.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORforever.html
remember it is only differentials:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DiffConst.html
remember it is only integrals:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Integral.html
remember it is only geometry:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SimpleEnough.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FullyAware.html
remember it is only transformations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroTransform.html
remember it is only calculations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Percentages.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FALSE.html
remember it is only groups:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroGroups.html
remember it is only logs:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LogsHuh.html
remember it is only vectors:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/IdiotVectors.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroVec.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/VectorLength.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/VectorSpaces.html
remember it is only polar coordinates:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PolarManager.html
remember it is only limits:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Limit.html
remember it is only equations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GOGI-GIGO.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Doofus.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroDistri.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ToothlessBite.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Competent.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/UseTrans.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Sheesh.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DivZero.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Think.html
remember it is only square roots:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GoodTeachers.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TwoTurds.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/STILL.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CanSpecify.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GrowUp.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Material.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Humour.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
remember it is only partial differential equations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff3.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff4.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NotFxy.html
Sorry - couldn't resist :-|
Dirk Vdm
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| User: "mL" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 06:46:56 PM |
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Dirk Van de moortel skrev:
"Androcles" <Engineer@hogwarts.physics_a> wrote in message
news:h0j7j.26873$kt3.19439@fe3.news.blueyonder.co.uk...
[snip]
Good luck with your mathematics but remember it is only
a description of reality, not reality itself.
remember it is only Boolean algebra:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
[..etc..]
Another one:
http://groups.google.co.uk/group/sci.physics/msg/cc5c16779a7fee3f
Androcles lecturing:
" I don't need to guess.
http://www.androcles01.pwp.blueyonder.co.uk/MC2.htm
Energy is very well defined in Newtonian Mechanics, it
is the integral of momentum with respect to time and momentum
is defined as mv. "
Is _it_ about integrals? or about Newtonian Mechanics?
/mel
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| User: "Dirk Van de moortel" |
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| Title: Re: Angular and linear velocity |
11 Dec 2007 10:52:51 AM |
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"mL" <mL.beyond@elsewhere.xxx> wrote in message news:46l7j.1479$R_4.1105@newsb.telia.net...
Dirk Van de moortel skrev:
"Androcles" <Engineer@hogwarts.physics_a> wrote in message news:h0j7j.26873$kt3.19439@fe3.news.blueyonder.co.uk...
[snip]
Good luck with your mathematics but remember it is only
a description of reality, not reality itself.
remember it is only Boolean algebra:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
[..etc..]
Another one:
http://groups.google.co.uk/group/sci.physics/msg/cc5c16779a7fee3f
Androcles lecturing:
" I don't need to guess.
http://www.androcles01.pwp.blueyonder.co.uk/MC2.htm
Energy is very well defined in Newtonian Mechanics, it
is the integral of momentum with respect to time and momentum
is defined as mv. "
Is _it_ about integrals? or about Newtonian Mechanics?
/mel
Seems that I overlooked that one :-)
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NoNeedToGuess.html
Thanks,
Dirk Vdm
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| User: "" |
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| Title: Re: Angular and linear velocity |
12 Dec 2007 05:50:47 AM |
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So can somebody plz tell me whether I was right?
Is it correct to say that the impulse has a magnitude of m * u and was
applied at b = 2/5 * a above the center of mass?
Thanks!
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| User: "Randy Poe" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 04:01:25 PM |
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On Dec 10, 4:49 pm, wrote:
Thanks for your answer.
It is said that the impulse is horizontal. You are right, I used a * P
because I assumed that the impulse was applied at the top. However,
we're not told so. So let's say that it is applied at a distance b
from the center of mass. Then b * P = 2/5*m*a^2 * w (that's because b
and P are perpendicular to each other). Is that correct?
Yes, that's OK.
Combining
with P = m * u from my first equation and using w = v * a
v = w*a
I get b =
2/5 * a^3 - is that sensible?
No, the units don't match up. You have length on the left
and length cubed on the right. That should tell you
something is wrong.
But fix the error above and you'll be OK.
And what is the answer to the question
(how big is the impulse)?
You already answered this. It's m*u.
- Randy
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| User: "" |
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| Title: Re: Angular and linear velocity |
10 Dec 2007 04:19:06 PM |
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Hi Randy,
v = w*a
Oups, of course!
No, the units don't match up. You have length on the left
and length cubed on the right. That should tell you
something is wrong.
But fix the error above and you'll be OK.
I now get that P = (2/5*m*a*u)/b - is this correct (I used w = u/a).
And what is the answer to the question
(how big is the impulse)?
You already answered this. It's m*u.
Right, but I also have my above expression. Setting both expressions
equal yields b = 2/5 * a. So is that my final answer? I apply an
impulse of magnitude m*u (or 2/5*m*a*u/b) at 2/5*a above the CM?
Again, thanks for your help.
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