Particle projected from above a dome

  • #1
Bling Fizikst
35
5
Homework Statement
refer to image
Relevant Equations
refer to image
Screenshot 2024-03-23 134640.png

Since the ball will follow a parabolic trajectory , i threw it into the coordinate plane with particle starting from the origin .

Screenshot 2024-03-23 142052.png

$$y= kx^2$$ Differentiating wrt ##t## : $$ gt=2ku^2 t\implies u=\sqrt{\frac{g}{2k}}$$ where $$k=\frac{\sin\theta}{R\cos^2\theta}$$ Now , final speed when it reached the bottom is $$\sqrt{u^2+2gh}$$ but I am stuck here on how to find ##h## in terms of ##\theta## in order to minimise the expression .
 
Physics news on Phys.org
  • #2
Are you sure you don't have to further analyse the collision with the sphere?
 
Last edited:
  • #3
Bling Fizikst said:
Now , final speed when it reached the bottom is $$\sqrt{u^2+2gh}$$ but I am stuck here on how to find ##h## in terms of ##\theta## in order to minimise the expression .
What you've shown is that the final speed is independent of the angle. That's a consequence of conservation of energy.
 
  • #4
PeroK said:
Are you sure you don't have to further analyse the collision with the sphere?
Maybe i misinterpreted the 'touching' of the dome . So , the correct interpretation should be that the ball collides with the dome and changes its trajectory and then strikes the ground ? I attached the figure . @Hill , yeah absolutely ! i was careless . It should be ##P(R\cos\theta ,h-R\sin\theta)##
 

Attachments

  • Screenshot 2024-03-23 161542.png
    Screenshot 2024-03-23 161542.png
    17.9 KB · Views: 9
Last edited:
  • #5
By the sketch, it appears that coordinates of the point P are ##(R \cos \theta, R \sin \theta)##. This is incorrect if the particle starts from the origin.
 
  • #6
Bling Fizikst said:
Maybe i misinterpreted the 'touching' of the dome . So , the correct interpretation should be that the ball collides with the dome and changes its trajectory and then strikes the ground ?
We can minimise the energy to ##mgh## by setting ##u \approx 0##. Why isn't that allowed? The reason must be that the ball will impact the dome more than once. Perhaps the constraint is that the ball must hit the dome once and only once?

That was my thinking
 
  • #7
Because of the post #5 above, the $$k=\frac{\sin\theta}{R\cos^2\theta}$$is incorrect.
 
  • #8
Bling Fizikst said:
@Hill , yeah absolutely ! i was careless . It should be ##P(R\cos\theta ,h-R\sin\theta)##.
Yes, but in fact you don't need ##\theta## at all.
 
  • #9
PeroK said:
We can minimise the energy to ##mgh## by setting ##u \approx 0##. Why isn't that allowed? The reason must be that the ball will impact the dome more than once. Perhaps the constraint is that the ball must hit the dome once and only once?

That was my thinking
I don't think this can be right. The problem gets too complicated. Instead, I think the question might assume that the ball is moving tangential to the sphere at the point of impact. That assumption leads to a solvable problem.

@Hill is that what you assumed?
 
  • Like
Likes phinds
  • #10
PeroK said:
I think the question might assume that the ball is moving tangential to the sphere at the point of impact. That assumption leads to a solvable problem.

@Hill is that what you assumed?
Yes, that is what I thought.
 
  • Like
Likes PeroK
  • #11
Was my original diagram correct then? If yes , after correction (thanks to @Hill) . I am getting $$\mid v_{\min}\mid =\sqrt{2gR(\sin\theta\pm \cos\theta)}$$ for ##\theta\in\left(0,\frac{\pi}{2}\right)## . Not sure how to proceed from here . The calculation looks tricky as i have to make cases for everything now. For instance , i took ##\sin\theta +\cos\theta \geq 1## and again you get different answers for taking ##\theta=0^{+}## or ##{\frac{\pi}{2}}^{-}## . All of them being wrong btw.
 
  • #12
Bling Fizikst said:
Was my original diagram correct then? If yes , after correction (thanks to @Hill) . I am getting $$\mid v_{\min}\mid =\sqrt{2gR(\sin\theta\pm \cos\theta)}$$ for ##\theta\in\left(0,\frac{\pi}{2}\right)## . Not sure how to proceed from here . The calculation looks tricky as i have to make cases for everything now. For instance , i took ##\sin\theta +\cos\theta \geq 1## and again you get different answers for taking ##\theta=0^{+}## or ##{\frac{\pi}{2}}^{-}## . All of them being wrong btw.
There is no need to involve theta. Note that both ##u## and ##h## are variables. You need a constraint on these to ensure that the ball impacts the sphere at a tangent.
 
  • #13
PeroK said:
There is no need to involve theta. Note that both ##u## and ##h## are variables. You need a constraint on these to ensure that the ball impacts the sphere at a tangent.
I dont know how to go about it .
 
  • #14
Bling Fizikst said:
I dont know how to go about it .
You have two curves. You need to calculate the condition so that the curves are tangent at their point of intersection.

Note that, in case you haven't already seen this, the parabolic path for an initially horizontal trajectory is:
$$y = h - \frac 1 2 gt^2, \ x = ut$$$$\implies y = h - \frac{gx^2}{2u^2}$$
 
  • #15
How about this?
Right the equation for the point ##P(x, kx^2)## to be at the distance ##R## from the dome center ##(0, h)##. You'll get a quadratic equation with solutions depending on ##k## and ##h##. Right a condition when this equation has exactly one solution, i.e., the parabola has one and only one point at the distance ##R## from the dome center. This will give you dependence of ##h## on ##k##. Then find ##k## that minimizes ##u##.

EDIT. Sorry, minimizes not ##u## but the speed at the ground.
 
  • #16
Hill said:
How about this?
Right the equation for the point ##P(x, kx^2)## to be at the distance ##R## from the dome center ##(0, h)##. You'll get a quadratic equation with solutions depending on ##k## and ##h##. Right a condition when this equation has exactly one solution, i.e., the parabola has one and only one point at the distance ##R## from the dome center. This will give you dependence of ##h## on ##k##. Then find ##k## that minimizes ##u##.
The calculation is brutal imo , i dont think it can be calculated by hand . We might have to look for other ways to get around this?
 
  • Sad
Likes PeroK
  • #17
Bling Fizikst said:
The calculation is brutal imo , i dont think it can be calculated by hand . We might have to look for other ways to get around this?
Hmmm... You might be right. I got, by hand, to ##h=\frac {1-4k^2R^2}{4k}##. (If my algebra is right.)
 
  • #18
Bling Fizikst said:
The calculation is brutal imo , i dont think it can be calculated by hand . We might have to look for other ways to get around this?
It's about a page of A4 without any shortcuts. You have to knuckle down and accept that these problems are not going to be cracked in a few lines.
 
  • Like
Likes Hill
  • #19
Hill said:
Hmmm... You might be right. I got, by hand, to ##h=\frac {1-4k^2R^2}{4k}##. (If my algebra is right.)
If ##k = \frac {g}{2u^2}##, then I have a plus in that equation.
 
  • Like
Likes Hill
  • #20
PeroK said:
You need a constraint on these to ensure that the ball impacts the sphere at a tangent.
Do we know for sure that this is the intent of the problem? I don't think it is worded very clearly.
 
  • #21
gmax137 said:
Do we know for sure that this is the intent of the problem?
We are pretty sure.
gmax137 said:
I don't think it is worded very clearly.
I agree. It should have been given as two steps: first find the relationship between ##h## and ##u## so that the ball touches the sphere at a tangent. Find ##h## and ##u## that minimises the speed of impact with the ground (which is equivalent to minimising the total energy of the ball (PE + KE)).
 
  • #22
Easily one of the worst problems for physics . Continuing from : $$(kx^2-h)^2+x^2=R^2$$ When two curves are tangent to each other at a point then there is a double root , here : ##(\alpha>0)## in it . Let ##\alpha,\alpha,\beta,\gamma## be the roots of the eqn : $$k^2x^4+(1-2hk)x^2+h^2-R^2=0$$ Define ##\sigma_r## as sum of roots taken ##r## at a time . $$\sigma_1=2\alpha+\beta+\gamma=0$$$$\sigma_4=\alpha^2 \beta\gamma=\frac{h^2-R^2}{k^2}$$ $$\sigma_3=\alpha^2(\beta+\gamma)+\beta\gamma(2\alpha)=0$$ Easily eliminating ##\beta,\gamma## in ##\sigma_3## to get : $$k^2\alpha^4=2(h^2-R^2)$$ Plugging the values of ##\alpha^4,\alpha^2## in the original equation to get :$$2k\sqrt{h^2-R^2}=2hk-1\implies h=\frac{4k^2 R^2+1}{4k}$$ Now consider the speed $$s=\sqrt{u^2+2gh}=\sqrt{2u^2+\frac{g^2 R^2}{u^2}}\geq \boxed{\sqrt{2Rg\sqrt{2}}}$$ equality at ##\boxed{u=\sqrt{\frac{gR}{\sqrt{2}}}}## $$\implies k=\frac{1}{R\sqrt{2}}\implies \boxed{h=\frac{3\sqrt{2}R}{4}}$$
 
Last edited:
  • Like
Likes Hill and PeroK
  • #23
Bling Fizikst said:
Easily one of the worst problems for physics . Continuing from : $$(kx^2-h)^2+x^2=R^2$$ When two curves are tangent to each other at a point then there is a double root , here : ##(\alpha>0)## in it . Let ##\alpha,\alpha,\beta,\gamma## be the roots of the eqn : $$k^2x^4+(1-2hk)x^2+h^2-R^2=0$$ Define ##\sigma_r## as sum of roots taken ##r## at a time . $$\sigma_1=2\alpha+\beta+\gamma=0$$$$\sigma_4=\alpha^2 \beta\gamma=\frac{h^2-R^2}{k^2}$$ $$\sigma_3=\alpha^2(\beta+\gamma)+\beta\gamma(2\alpha)=0$$ Easily eliminating ##\beta,\gamma## in ##\sigma_3## to get : $$k^2\alpha^4=2(h^2-R^2)$$ Plugging the values of ##\alpha^4,\alpha^2## in the original equation to get :$$2k\sqrt{h^2-R^2}=2hk-1\implies h=\frac{4k^2 R^2+1}{4k}$$ Now consider the speed $$s=\sqrt{u^2+2gh}=\sqrt{2u^2+\frac{g^2 R^2}{u^2}}\geq \boxed{Rg\sqrt{2}}$$ equality at ##\boxed{u=\sqrt{\frac{gR}{\sqrt{2}}}}## $$\implies k=\frac{1}{R\sqrt{2}}\implies \boxed{h=\frac{3\sqrt{2}R}{4}}$$
##R## and ##g## have numerical values in the problem.
 
  • #24
I had another idea to use vectors. I took the origin at the centre of the sphere. The particle's position and velocity are given by:
$$\vec x = (x, h - \frac{gx^2}{2u^2}), \ \vec v = (u, - \frac{gx} u)$$If the particle impacts the sphere at a tangent, then:
$$\vec x = (R\cos \theta, R \sin\theta) \ \text{and} \ \vec x \cdot \vec v = 0$$$$\vec x \cdot \vec v = uR\cos \theta - \frac{R^2g}{u}\sin \theta\cos \theta = 0$$$$\implies R\sin \theta = \frac{u^2}{g}$$$$\implies y = \frac{u^2}{g}, x^2 = R^2 - \frac{u^4}{g^2}$$This allows us to calculate ##h##:
$$h - \frac{gx^2}{2u^2} = R \sin\theta = \frac{u^2}{g}$$$$\implies h = \frac{u^2}{2g} + \frac{R^2g}{2u^2}$$
PS note that when the energy is minimised, with ##u^2 = \frac{gR}{\sqrt 2}##, then:
$$\sin \theta = \frac{u^2}{Rg} = \frac 1 {\sqrt 2}$$And ##\theta = \dfrac \pi 4##.
 
  • Like
Likes Bling Fizikst
  • #25
Since the OP has posted a solution, here is a YouTube video with a different approach, if of interest.
 
  • Like
Likes TSny
  • #26
Steve4Physics said:
Since the OP has posted a solution, here is a YouTube video with a different approach, if of interest.
Yes, that was slick.

At time 2:40 he uses "AM greater than or equal to GM", which didn't ring a bell. After some thought it occurred to me that AM refers to arithmetic mean and GM to geometric mean. See AM-GM inequality. That was a clever way to find the minimum of ##2\cos \theta + \dfrac 1 {\cos \theta}## :oldsurprised:
 
  • #27
TSny said:
Yes, that was slick.
I think the simplest solution is to look at the distance squared, ##D##, of the particle from the centre of the sphere as a function of ##y##:
$$y = h - \frac{gx^2}{2u^2} \ \implies x^2 = (h-y)\frac{2u^2}{g}$$
$$D(y) = x^2 + y^2 = (h-y)\frac{2u^2}{g} + y^2$$$$D'(y) = 2y - \frac{2u^2}{g}$$Then:
$$D'(y) = 0 \implies y = \frac{u^2}{g}$$And:
$$D_{min} = \frac{2u^2}{g}h - \frac{u^4}{g^2}$$Finally:
$$D_{min} = R^2 \implies h = \frac{R^2g}{2u^2} + \frac{u^2}{2g}$$The final speed satisfies:
$$v^2 = u^2 + 2gh = 2u^2 + \frac{g^2R^2}{u^2}$$To minimise this is equivalent to minimising the function:
$$f(z) = az +\frac b z$$$$f'(z) = 0 \implies a - \frac b {z^2} = 0 \implies z^2 = \frac b a$$This gives a minimum when:
$$u^4 = \frac{g^2R^2}{2}$$$$u^2 = \frac{gR}{\sqrt 2}$$And:
$$v^2_{min} = 2\sqrt 2 gR$$
 
  • #28
PeroK said:
I think the simplest solution is to look at the distance squared, ##D##, of the particle from the centre of the sphere as a function of ##y##:
Yes, that's a very nice approach!

When you got to the following part, I couldn't resist finding the minimum of ##f(z)## using the "AM-GM Inequality":

PeroK said:
To minimise this is equivalent to minimising the function:
$$f(z) = az +\frac b z$$
AM ##= \frac 1 2 (az + \dfrac b z) = \frac 1 2 f(z) ## and GM ## = \sqrt{ab}##. So, ##f_{min} = 2 \sqrt{ab} = 2 \sqrt 2 gR##.
 
  • #29
TSny said:
AM ##= \frac 1 2 (az + \dfrac b z) = \frac 1 2 f(z) ## and GM ## = \sqrt{ab}##. So, ##f_{min} = 2 \sqrt{ab} = 2 \sqrt 2 gR##.
How do you justify that in this case AM = GM? In general, AM > GM is possible.

PS We have equality if ##az= \frac b z##. Interesting.
 
  • #30
I think you have to be careful with that technique. In general, the minimum of the function ##f(z) = g(z) + h(z)## occurs when ##g'(z) + h'(z) = 0## and not when ##g(z) = h(z)##. The GM of ##\sqrt{g(z)h(z)}## may be unrelated to the minimum value.
 
  • #31
PeroK said:
I think you have to be careful with that technique. In general, the minimum of the function ##f(z) = g(z) + h(z)## occurs when ##g'(z) + h'(z) = 0## and not when ##g(z) = h(z)##. The GM of ##\sqrt{g(z)h(z)}## may be unrelated to the minimum value.
Yes, but in our case the GM is independent of ##z##.

In our case, ##x = az## and ##y = b/z## with ##a## and ##b## positive numbers. So, the GM of ##x## and ##y## is GM = ##\sqrt{ab}## for all nonzero values of ##z##. Hence the AM of ##x## and ##y## must be ##\ge \sqrt{ab}## for all ##z##. $$ \text{AM} = \frac 1 2 (x + y) = \frac 1 2 (az + \frac b z) \ge \sqrt{ab}$$To see that the minimum value of the AM is, in fact, ##\sqrt{ab}##, we note that it should be clear that there is a value of ##z## such that ##az = b/z## and, hence, ##x = y##. When ##x = y##, the AM of ##x## and ##y## equals the GM of ##x## and ##y##. So, we can conclude that the minimum value of ##\frac 1 2 (az + b/z) = \sqrt{ab}##.

I'm sure that's not the clearest argument. But I hope it makes sense.
 
  • #32
TSny said:
Yes, but in our case the GM is independent of ##z##.
The technique works as long as the minimum occurs when the two functions are equal. Which is the case here. But, in general, it is not a valid technique for minimizing a function.
 
  • #33
I agree, it's not a general method of finding a minimum. But, I think it's neat when it applies (which probably isn't that often).

Take ##F(x) = \dfrac{8 \cot x}{\sqrt x} + 2 \sqrt x \tan x##

The product of the two terms is the constant 16. The first term monotonically decreases for ##0 < x < \pi/2## and the second term monotonically increases for ##0< x < \pi/2##. So, there must be a value of ##x## in this range where the two terms are equal. So, we can use the AM-GM inequality to see that the minimum of ##F(x)## in this range is ##2\sqrt{16} = 8##.
 
  • #34
TSny said:
I agree, it's not a general method of finding a minimum. But, I think it's neat when it applies (which probably isn't that often).

Take ##F(x) = \dfrac{8 \cot x}{\sqrt x} + 2 \sqrt x \tan x##
It applies to any function of the form ##af(x) + \frac b {f(x)}## , which that one is.
 
  • #35
PeroK said:
It applies to any function of the form ##af(x) + \frac b {f(x)}## , which that one is.
Yes.

Edit: I think there is also the condition that the two terms ##af(x)## and ##\dfrac b {f(x)}## be positive.
 
Last edited:

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
602
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
766
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
788
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
777
  • Introductory Physics Homework Help
Replies
31
Views
4K
Back
Top