General Motion of a particle in 3 dimensions

[YauYauYau]

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.

 

[haruspex]

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.

 

[YauYauYau]

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

 

[ehild]

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. :)

 

[haruspex]

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?

 

[YauYauYau]

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.