Projectile motion on a hemisphere

[Bread18]

Homework Statement

A person standing on the top of a hemispherical rock of radius R kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity $v_i$
What must be it's minimum initial speed if the ball is never to hit the rock after it is kicked?

Homework Equations

$v=v_i+at,\\ v^2=v_i ^2 + 2ar, \\ r=v_i t +\frac{1}{2}at^2$

The Attempt at a Solution

I'm not sure how to do this one as a parabola and a semicircle are 2 different shaped curves, a little nudge in the right direction would be helpful, thanks guys.

 

[tiny-tim]

Hi Bread18!

… a parabola and a semicircle are 2 different shaped curves, a little nudge in the right direction would be helpful, thanks guys.

Write out the two equations, and see where they intersect …

what do you get? ​

 

[Bread18]

Hmm ok...
Well the eqn of the semicircle is $y=\sqrt{R^2 - x^2}$

The eqn of motion is $y = \frac{1}{2}at^2 \\ x = v_it \\ y = \frac{1}{2}a(\frac{x}{v_i})^2 \\ 2yv_i^2 = a x^2?$

 

[tiny-tim]

$2yv_i^2 = a x^2?$

= a(R² - y²) ?

 

[Bread18]

$ay^2 + 2yv_i^2 - aR^2 = 0 \\ 4v_i^4 +4a^2R^2 < 0 \\ v_i^4 < -a^2R^2$

and that's not right...

 

[tiny-tim]

$ay^2 + 2yv_i^2 - aR^2 = 0$

after that, you've lost me

 

[Bread18]

Solve the quadratic, they don't touch so the discriminate needs to be < 0 (typo in the other post, I'll fix it)

 

[tiny-tim]

ah, sorry, i'd forgotten what the question asked for! …

What must be it's minimum initial speed if the ball is never to hit the rock after it is kicked?

yes, that's fine …

$ay^2 + 2yv_i^2 - aR^2 = 0 \\ 4v_i^4 +4a^2R^2 < 0 \\ v_i^4 < -a^2R^2$

remember, your "a" was negative!

 

[Bread18]

Yes but it's a^2, so it cancels out the negative, leaving me with -g^2R^2?

 

[tiny-tim]

hmm … you're right!

ok, let's go back and check your original equations …

Well the eqn of the semicircle is $y=\sqrt{R^2 - x^2}$

The eqn of motion is $y = \frac{1}{2}at^2 \\ x = v_it \\ y = \frac{1}{2}a(\frac{x}{v_i})^2 \\ 2yv_i^2 = a x^2?$

ah! should be y = R + 1/2 at² !

(no wonder it couldn't avoid hitting the circle! )

 

[Bread18]

hmm … you're right!

ok, let's go back and check your original equations …


ah! should be y = R + 1/2 at² !

(no wonder it couldn't avoid hitting the circle! )

Haha well spotted

So, now with that fix, I get $2v_i^2(y-R)=ax^2 \\ 2v_i^2(y-R)=a(R^2 - y^2) \\ 2v_i^2 = -a(R+y) \\ v_i^2 = \frac{g}{2}(R+y)$

 

[tiny-tim]

looks good!

(but on my recent performance, i could be missing something! )

 

[Bread18]

looks good!

(but on my recent performance, i could be missing something! )

Haha yeah, we've all been missing simple things..
I don't see how this ties in with it not hitting the semi circle though.

 

[tiny-tim]

because …

$v_i^2 = \frac{g}{2}(R+y)$

gives you the y-coordinate where it hits the circle

for example, if v = 0, the ball drops straight down, and hits the circle at y = -R !

ok, now what happens to y as you increase v ? ​

(btw, the other solution, y = R for any value of v was eliminated from the equation when you divided it by (y - R) )

 

[Bread18]

So I want y = 0?

 

[tiny-tim]

in a moment, I'm going out for an hour or so

draw yourself some diagrams of a parabola touching and cutting a circle, and see what happens as the parabola changes shape

(also, think how many times can a parabola touch and cut a circle?)

 

[Bread18]

Ok, thanks for your help, I think I'm going to go to bed now (2am here). Hopefully When I wake up it'll all become obvious...