Projectile Motion: Finding Rock's Min. Speed & Ball's Landing Point

[blackboy]

Homework Statement

A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest) to give it a horizontal velocity vi.
a. What must be the rock’s minimum speed if the ball is never to hit the rock after it is kicked?
b. With this initial speed, how far from the base of the rock does the ball hit the ground?

Homework Equations

Projectile Motion

The Attempt at a Solution

I tried my way a couple times but the answer I get is wrong. Ok I tried to find the function of the curve of the rock. Setting y=0 at the top of the rock, I get y=-(gt²)/2. And vt=R. Solving for t when y=-R t=√(2r/g) and v=(√Rg)/2. Is this not vmin? I even checked it by putting vmin back into the y as function of x. Then I put something bigger back into that same function and y is greater for vmin. This means that the ball is always higher than the rock if kicked greater than (√Rg)/2. The answer key says vmin=(√Rg). What did I do wrong?

 

[LowlyPion]

To not touch the rock doesn't the initial velocity need to satisfy

m*v²/R > m*g

Armed with that velocity which is horizontal, and the time to fall determined by R, then you know how far away.

 

[blackboy]

Um don't know about m yet. Can you explain what I did wrong?

 

[LowlyPion]

Um don't know about m yet. Can you explain what I did wrong?

m cancels out if you might notice.

That expresses the contact relationship, which is that it will lose contact when the centripetal acceleration is greater than gravity.

 

[RoyalCat]

Spoiler

LowlyPion, losing contact isn't the question here, the ball is kicked and loses contact at t=0
Forces are completely irrelevant. Though your approach looks interesting, I can't quite make heads or tails of it.

Ooh, I see now. My approach is much more time-consuming and yours gives the solution straight away. @@

The approach I'd choose here would be kinematics and analytic geometry.
Do you know how to construct the y=f(x) equation representing a ballistic trajectory, and the y=f(x) equation representing a hemisphere?

The latter requires a tiny bit of knowledge to develop on your own, so if you get stuck, here's a spoiler, but knowing how to develop the former is critical to all such problems.

Spoiler

The equation representing a circle of radius R is:
R²=x²+y²
Can you transform this into a function of the form y=f(x)?

 

[LowlyPion]

LowlyPion, losing contact isn't the question here, the ball is kicked and loses contact at t=0

Consider a ball on a string that is cut exactly at the top of the sphere. If the speed v at t=0 meets the condition that v²/R > g, then it will not touch the sphere now will it. If the speed is less than that, then were it to continue, since the tension is not sufficient to keep it taught, it would necessarily come in contact, in fact be in contact, with the sphere at the top.

 

[RoyalCat]

Consider a ball on a string that is cut exactly at the top of the sphere. If the speed v at t=0 meets the condition that v²/R > g, then it will not touch the sphere now will it. If the speed is less than that, then were it to continue, since the tension is not sufficient to keep it taught, it would necessarily come in contact, in fact be in contact, with the sphere at the top.

Heh, 100% correct, the exact result my geometry got me. Could you please elaborate on why this is correct, please?

 

[Redbelly98]

m*v²/R > m*g

LowlyPion, losing contact isn't the question here, the ball is kicked and loses contact at t=0
Forces are completely irrelevant. Though your approach looks interesting, I can't quite make heads or tails of it.

LowlyPion is using the centripetal force for circular motion:

[The force on a mass constrained to move on the sphere's surface at constant velocity v]​

is greater than

[The actual force mg]​

Note: if the OP is now learning about uniformly accelerated motion, then centripetal force & acceleration has probably not been covered in class yet.

 

[Redbelly98]

The Attempt at a Solution

I tried my way a couple times but the answer I get is wrong. Ok I tried to find the function of the curve of the rock. Setting y=0 at the top of the rock, I get y=-(gt²)/2. And vt=R. Solving for t when y=-R t=√(2r/g) and v=(√Rg)/2. Is this not vmin? I even checked it by putting vmin back into the y as function of x. Then I put something bigger back into that same function and y is greater for vmin. This means that the ball is always higher than the rock if kicked greater than (√Rg)/2. The answer key says vmin=(√Rg). What did I do wrong?

When x=vt=R, y is not -R.

That would be true if the ball were constrained to move along the sphere's surface, but it does not do that. It will be above the surface, so y is somewhere between 0 and -R when x=R.

 

[Phrak]

Homework Statement

A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest) to give it a horizontal velocity vi.
a. What must be the rock’s minimum speed if the ball is never to hit the rock after it is kicked?
b. With this initial speed, how far from the base of the rock does the ball hit the ground?

...

 

[LowlyPion]

Note: if the OP is now learning about uniformly accelerated motion, then centripetal force & acceleration has probably not been covered in class yet.

As a side note, this is another problem in the series he was dealing with earlier.
https://www.physicsforums.com/showthread.php?t=320363

I should hope he is learning about centripetal acceleration by now.

 

[LowlyPion]

...

That can only be a typo.

 

[Redbelly98]

As a side note, this is another problem in the series he was dealing with earlier.
https://www.physicsforums.com/showthread.php?t=320363

I should hope he is learning about centripetal acceleration by now.

Okay, yes, that is clearly a centripetal acceleration problem so it has been covered.

 

[RoyalCat]

LowlyPion is using the centripetal force for circular motion:

[The force on a mass constrained to move on the sphere's surface at constant velocity v]​

is greater than

[The actual force mg]​

Note: if the OP is now learning about uniformly accelerated motion, then centripetal force & acceleration has probably not been covered in class yet.

I can see how when:
v²/R > g
it follows that the body loses contact with the hemisphere (From the accelerated frame, the net force points up).

But how does that show that it won't return to the hemisphere? I know that v²/R > g is the correct answer, from my own calculations, but I don't quite see how the above explanation of losing contact works for the rest of the motion.

Thanks Redbelly98, I'll have to read up on curvatures to see what you mean, exactly. Hopefully the wiki article will be enough. :s

 

[Redbelly98]

For a parabola, the path's curvature gets straighter with time. Whereas the circular path maintains a constant curvature.

If the curvatures were equal initially, the parabola always has less curvature for t>0. Therefore it can never meet the circle again.

 

[LowlyPion]

For any object traveling along such a hemisphere, g=v²/R, so any object that fulfills v²/R > g, is necessarily not traveling along the hemisphere, right?

The first part is not necessarily true. g = v²/R is merely the condition under which the ball exerts at most 0 contact force on the surface at any point around the sphere. The top of course is where gravity acts normally into the surface, so like New York, NY, if it won't touch it there, it won't touch it anywhere.

 

[blackboy]

I am sorry I am not following along with any of the force or mass stuff. I know I am close but what did I do wrong?

 

[Phrak]

That can only be a typo.

I prefer the problem where the rock is moving.

 

[blackboy]

Ok you if you want the rock to be moving go ahead. Still what did I do wrong? I am going crazy over this problem!

 

[ideasrule]

"And vt=R. Solving for t when y=-R t=√(2r/g) and v=(√Rg)/2. Is this not vmin?"

That's the problem: by using vt=R, you're assuming that at vmin, the ball touches the hemisphere when it gets to the ground. It can't; it must land a certain distance away from the hemisphere, and part b) of your question asks you to find this distance.

 

[Redbelly98]

I am sorry I am not following along with any of the force or mass stuff. I know I am close but what did I do wrong?

See post #9. Feel free to ask questions about what I said there.

 

[blackboy]

I started over. I got x²+y²=R². Now the ball always has to be over this right,so I got
y²> R²-x². For the ball's y position I got y=R-gt²/2. Then I realized vt=x so t=x/v. So
y=R-(gx²)/(2v²). Substituting in and simplifying I got (gxⁿ)/(4vⁿ)+x²>(Rgx²)/(v²) where n=4. Dividing by x² I finally got (gx²)/(4vⁿ)+1>(Rg)/(v²) where n=4. What now, I am undoubtedly close but that x has to get in the way? Thanks.

 

[Redbelly98]

Awesome, you are nearly done!

I started over. I got x²+y²=R². Now the ball always has to be over this right,so I got
y²> R²-x². For the ball's y position I got y=R-gt²/2. Then I realized vt=x so t=x/v. So
y=R-(gx²)/(2v²). Substituting in and simplifying I got (gxⁿ)/(4vⁿ)+x²>(Rgx²)/(v²) where n=4. Dividing by x² I finally got (gx²)/(4vⁿ)+1>(Rg)/(v²) where n=4. What now, I am undoubtedly close but that x has to get in the way? Thanks.

Since x² must be positive for x≠0, you're inequality will hold as long as

1 > Rg / v²​

Take it from there . . .

 

[blackboy]

I know x² will be positive but what gives us the right to get rid of the expression? Sorry.

 

[Redbelly98]

Fair question.

We know that, for x≠0,

(gx²)/(4v⁴)+1 > 1​

So, as long as

1 > Rg / v²​

the inequality must hold true.

p.s. FYI, you can type "[noparse]x⁴[/noparse]" → x⁴ to express any exponent or superscript.

 

[RoyalCat]

I started over. I got x²+y²=R². Now the ball always has to be over this right,so I got
y²> R²-x². For the ball's y position I got y=R-gt²/2. Then I realized vt=x so t=x/v. So
y=R-(gx²)/(2v²). Substituting in and simplifying I got (gxⁿ)/(4vⁿ)+x²>(Rgx²)/(v²) where n=4. Dividing by x² I finally got (gx²)/(4vⁿ)+1>(Rg)/(v²) where n=4. What now, I am undoubtedly close but that x has to get in the way? Thanks.

The approach presented in this post is exactly the same, but avoids the conflict you got by having x² somehow left in there.

y(rock) = √(R²-x²)
This is not true for a full circle, but is true for a hemisphere, the circle equation would have a ± sign before the square root, but would not be a function.

y(ball) as a function of t = R-½gt²
x(ball) as a function of t = vt
t(ball) as a function of x = x/v

y(ball) as a function of x = R-(g/2v²) * x²

For our solution, we want the two graphs to intersect at one point, and one point only. That is, where x=0.

To make sure this happens:

y(ball) = y(rock)
√(R²-x²) = R-(g/2v²)x² (We can square both sides, as they are both >0 for any x that we're interested in (We don't care about the ball continuing through the floor))

R²-x² = (R - (g/2v²)x²)²
R²-x² = R² - (gR/v²)x² + (g²/2v⁴)x⁴
(1-gR/v²)x² + (g²/2v⁴)x⁴ = 0
From here we can extract the one solution we're interested in, x=0, by dividing both sides by x² we get:
(1-gR/v²) + (g²/2v⁴)x² = 0
This is a quadratic equation, and we want to demand that it has no solutions.

discriminant < 0 is the appropriate condition
0²-4*(1-gR/v²)(g²/2v⁴) < 0
(((v²-gR)/v²))(g²/2v⁴) > 0

(g²/2v⁴) > 0 this is a squared expression, therefore dividing by it will not change the sign of the inequality.

((v²-gR)/v²) > 0
-->
v²/R > g

You're welcome, glad I could help. :)
Just use the advanced posting feature to look at all the nifty things you can do with text here.
If I could only be bothered to learn LaTeX. :(

 

[blackboy]

p.s. FYI, you can type "[noparse]x⁴[/noparse]" → x⁴ to express any exponent or superscript.

I don't know the codes, I got the things from your copy/paste math symbols.

OK I got what you said. Thank you a lot RedBelly98! Also thank you RoyalCat for your post, for it was a better way than mine! I can get the second part now.