prove velocity of a ball

Question

A solid spherical ball which rests in equilibrium at the interior bottom of a fixed spherical globe is perfectly

rough. the ball is struck a horizontal blow of such magnitude that the initial speed of its centre is v. Prove

that, if v lies between (10 dg/7)^1/2 and (27 dg/7)^1/2, the ball will leave the globe, d being the difference

between the radii of the ball and globe.

Answer 1

I'm assuming that when you say leave the globe, you mean that it loses contact with the inside of the globe, as the problem seems to imply that the ball is completely enclosed within the globe.  There are two cases to consider.  First, if the ball has the energy only to make it to the midpoint of the globe (where the slope is vertical), then it will roll back down without leaving the side.  To calculate this, we need only to equate the initial energy at the bottom with what the energy would be at this point.

 

So, initially, we have kinetic energy 1/2 mv² and rotational kinetic energy 1/2 (2mr²/5)(v/r)².  At the turn around, it has only potential energy mgd.  Therefore,

 

1/2 mv² + mv²/5 = mgd

7 v² / 10 = gd

v = √(10gd/7)

 

Anything above this will depart the surface of the globe (fall back toward the center) unless it also has enough energy such that centripetal acceleration is at least equal to gravity at the top.  If this condition is met at the top, then at the top,

 

v_t²/d = g

v_t = √(dg)

 

So, equation our initial and final energies, we have

 

1/2 mv₀² + mv₀²/5 = 1/2 mv_t² + mv_t²/5 + mg(2d)

 

7 v₀² / 10 = 7 dg / 10 + 2gd

 

v₀² = 27 dg / 7

v_⁰ = √(27dg/7)

 

Anything in between these will fall off the globe without completing the loop.

 

If you have any additional questions on this, please let me know.

Answer 2

Hello Samagra,

 

Before we do any work on this problem, let's sit back and think about the physical situation.

 

We've got a ball sitting at the bottom of a spherical shell, and we're going to whack the ball so that it starts rolling around the inside. What could happen?

 

1) We could hit the ball so lightly that it doesn't even roll past the halfway point going up the side of the shell. The ball will roll for a little bit up the side, slow down, reverse direction, and roll right back down at us. In this case, the ball is always in contact with the shell.

 

2) We could hit the ball so hard that it rolls all the way to the top of the shell and keeps rolling down the other side. The ball will complete a full loop inside the shell. In this case, the ball is always in contact with the shell.

 

3) We could hit the ball hard enough to make it past the halfway point going up the side of the shell, but not hard enough for it to make it across the top of the shell. In this case, the ball will roll up the shell, lose speed, and fall down through thin air. In this case, the ball is no longer in contact with the shell.

 

Okay, so we've figured out how the ball will leave the shell. Our job is to find the minimum initial velocity of the ball so that it can pass the halfway point going up the side of the shell. We also need to find the maximum initial velocity of the ball so that it can't complete a loop.

 

That was the brainy part of the problem. Hopefully, that's enough to get you started on solving the problem! :)

 

 

 

If not, use these hints...

 

Hint #1: The minimum initial velocity of the ball happens when we hit the ball just hard enough so that it rolls exactly halfway up the side of the shell. Find the initial energy of the ball at the bottom of the shell, just after we've hit it. Find the final energy of the ball at the halfway point up the side of the shell. Use the Law of Conservation of Energy.

 

 

 

 

 

Hint #2: The maximum initial velocity of the ball happens when we hit the ball hard enough so that the centripetal force exerted by the shell on the ball is exactly equal to the gravitational force on the ball while the ball is at the top of the shell. Find the initial and final energies of the ball, then use Conservation of Energy.

 

 

 

 

 

Hint #3: Total Energy = Gravitational P.E. + Translational K.E. + Rotational K.E.

 

 

 

 

 

Hint #4: Gravitational P.E. = mgh, Translational K.E. = ½mv², and Rotational K.E. = ½Iω²

 

 

 

 

 

Hint #5: The moment of inertia for a solid sphere: I = (2/5)mr²

 

 

 

 

 

Hint #6: Angular velocity is defined: ω = v/r, and the problem defines d = R - r

 

 

 

 

 

 

Hint #7: Centripetal force: F_centripetal = mv²/r_centripetal, where r_centripetal is the radius of the curved path the ball's center of gravity is taking.

 

 

 

 

Best of luck! :)