Minimum Initial Speed for Ball to Clear Hemispherical Rock

[nebullient]

Homework Statement

A person standing at 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 its minimum initial speed if the ball is never to hit the rock after it is kicked?

Homework Equations

The Attempt at a Solution

$d_y = 1/2at^2$
$-R = -1/2gt^2$
$2R = gt^2$
$t = { \sqrt \frac {2R} {g}}$

$d_x = v_xt$
$d_x = v_it$
$v_i = \frac {d_x} { \sqrt {\frac {2R} {g}}}$ If $d_x = R$, then

$v_i = \frac {R} { \sqrt {\frac {2R} {g}}} = { \sqrt {\frac {Rg} {2}}}$ But since $d_x > R$, $v_i > { \sqrt {\frac {Rg} {2}}}$However, the answer is just $v_i > { \sqrt {Rg} }$Please let me know where I went wrong.

 

[PumpkinCougar95]

Your assumption that in order for it to not strike again, its range needs to be bigger than $R$ is wrong. What other constraints can you think of?

HINT: Distance from Center of hemisphere is always greater than R

EDIT: Its range would be greater than R but that's not the only constraint.

 

[nebullient]

But the horizontal displacement must be greater than R for the ball not to hit the rock...

 

[PumpkinCougar95]

Yes it needs to be, But there is another constraint.

 

[nebullient]

I can't think of anything else :( Is it that the vertical displacement must be greater than R? I used $d_y = -R$ because the ball is initially at rest and it falls to the ground.

 

[PumpkinCougar95]

Write the Coordinates as functions of time first.

 

[PumpkinCougar95]

Did you get these ?

$$ x= v_i t$$
and
$$ y = R - \frac {gt^2} {2} $$

 

[nebullient]

Yes, but I wrote the second equation differently.

 

[PumpkinCougar95]

Oh, Did you have taken the top point as origin?

Now, What is the distance between the centre of the hemisphere and the ball at some random time?

 

[nebullient]

$\sqrt {x^2 + y^2}$

 

[Gene Naden]

Hi,
I initially got the same answer as nebullient, but then I got to thinking about the ball when it is close to the point where it was kicked. If $v_i < { \sqrt {Rg} }$ then the ball never gets off of the sphere. I used a coordinate system with the origin at the center of the sphere and calculated x and y for the sphere, as well as the ball, and set them equal to each other. This led me to the relation $\frac{x^2}{R} = g \frac{x^2}{v_i^2}$, which yields $v_i > { \sqrt {Rg} }$

 

[nebullient]

If vi<√Rgvi<Rgv_i < { \sqrt {Rg} } then the ball never gets off of the sphere.

How did you determine this?

 

[PumpkinCougar95]

x2+y2

Good, Now just put in the values and then you will get ( $\sqrt{x^2 + y^2} > R$)

$$ \frac {g^2t^2}{4} >= Rg - v_i ^2$$

So $Rg- v_i^2$ must be negative.

 

[Gene Naden]

I wrote x and y as a function of t, for the ball. I wrote y as a function of x for the surface of the sphere and used the approximation $\sqrt{(R^2-x^2)}=R(1-\frac{1}{2}\frac{x^2}{R^2})$. I substituted for t as a function of $v_i$ and x. Setting the two expressions for y equal to one another gave the result. This is just the speed you need to get off of the surface of the sphere.

If $v_i = { \sqrt {\frac {Rg} {2}}}$ then the sphere gets in the way of the ball.

Good luck!

 

[haruspex]

The easiest way is to think about curvature. The initial trajectory must not curve more tightly than the sphere. The trajectory's curvature is a result of the centripetal force provided by gravity.

 

[Gene Naden]

That sounds like a less roundabout way to approach the solution.