General Motion of a particle in 3 dimensions

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  • #1
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Homework Statement


Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the
wheel is v0, and the radius of the wheel is b, show that the greatest height above the ground that the mud can go is
b + v02 / 2g + gb2/ 2v02
At what point on the rolling wheel does this mud leave?
(Note: It is necessary to assume that v02≥bg.)

Homework Equations


In 3 dimensions concept,
r = i b cosθ + j b sinθ
Since v = rω = rθ',
v = dr / dt = (-b sin θ i + b cos θ j) θ' = -v0 ( -sin θ i + cos θ j )

The Attempt at a Solution


Firstly, find the time when the particle meets maximum height.
Take downwards as positive
for vertical direction,
v = u + at
0 = v0 cos θ + g t
Then I find the time with minus sign.
It sounds quite weird.
In case, how do I know which side of rim the particle is thrown from rolling wheel?
The forum has already posted that problem already but still I have no idea with it.
Thanks.
 

Answers and Replies

  • #2
haruspex
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v = dr / dt = (-b sin θ i + b cos θ j) θ' = -v0 ( -sin θ i + cos θ j )
Not sure how you intend this equation. What is r here?
I think using this i and j notation is of no benefit. I would just consider a mud particle released at some point on the wheel (some theta) and find its vertical height and velocity.
 
  • #3
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Not sure how you intend this equation. What is r here?
I think using this i and j notation is of no benefit. I would just consider a mud particle released at some point on the wheel (some theta) and find its vertical height and velocity.
I just try to transform it into polar coordinate by r = ix + my
Don't know whether it is correct or not
 
  • #4
ehild
Homework Helper
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Homework Statement


Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the
wheel is v0, and the radius of the wheel is b, show that the greatest height above the ground that the mud can go is
b + v02 / 2g + gb2/ 2v02
At what point on the rolling wheel does this mud leave?
(Note: It is necessary to assume that v02≥bg.)

Homework Equations


In 3 dimensions concept,
r = i b cosθ + j b sinθ
Since v = rω = rθ',
v = dr / dt = (-b sin θ i + b cos θ j) θ' = -v0 ( -sin θ i + cos θ j )
why is it 3 dimension? The particle and the wheel move in a plane. Why do you have the minus sign in front of v?
I think you mean r the position vector with respect to the centre of the wheel. So you treat the problem in the frame of reference of the wheel. And θ means the angle of r with respect to the positive horizontal axis. You also consider y positive upward. You should make the difference between a vector and its magnitude, denoting a vector by bold letter, for example.
So r = i b cosθ + j b sinθ. b=r here.
v = dr / dt = (-b sin θ i + b cos θ j) θ' = v0 ( -sin θ i + cos θ j )

The Attempt at a Solution


Firstly, find the time when the particle meets maximum height.
Take downwards as positive
for vertical direction,
v = u + at
0 = v0 cos θ + g t
If upward is positive you should take g with negative sign.
The particle does not start from zero height. Its height is b sinθ with respect to the centre of the wheel, when it leaves the wheel.

In case, how do I know which side of rim the particle is thrown from rolling wheel?
The forum has already posted that problem already but still I have no idea with it.
Thanks.
Find the maximum height at a given angle; then find the angle which corresponds to the maximum of the maximum height. :)
 
  • #5
haruspex
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I just try to transform it into polar coordinate by r = ix + my
Don't know whether it is correct or not
Yes, but what does r represent? A point on the rim? The position of a mud particle that has been thrown up? And where is the origin? Is it a fixed point on the ground?
Please, make it easier on everyone and consider a particle thrown from a point on the wheel at some angle theta around the wheel from, say, the top. What is its launch angle and speed?
 
  • #6
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Anyway thanks. I got the idea finally.
In the beginning, I misunderstand where the particle goes ( in case, now I know it is just assumed by me )
Then I realize how the coordinate comes.
 

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