M3 4F question 12

tez_9

wondered if any1 has done this, its a real mother of a question

Reply 1

BCHL85

what's it?

Reply 2

Christophicus

tez_9

wondered if any1 has done this, its a real mother of a question

I'm very near the end of this chapter. Once i've finished the work, i'll have a go. Not quite there yet though!

Reply 3

tez_9

A bowl is made from a smooth spherical shell of radius r by cutting away the part which is more than 1/3r above the horizontal plane through its centre. A marble is projected from the lowest point of the bowl with speed u. Show that in the subsequent motion the marble will leave the bowl and not fall back into it provided u^2 > 17gr/3

Reply 4

dvs

Using pythagoras you can find the length (x) of the opening in the shell:
(x/2)² + (r/3)² = r²
x = (4/3)sqrt[2] r

Since the marble was performing circular motion, it initially leaves the shell with a path that is tangent to the surface of the shell. In other words, it becomes a projectile with an angle A to the horizontal. Since its path is tangent to the circle, A=90-B, where B is the angle the radius makes with the point at the edge of the opening.

Basic trigonometry should give you:
cosB = (r/3)/r = 1/3
sinB = (x/2)/r = {[(4/3)sqrt[2] r]/2}/r = (2/3)sqrt[2]
(Hopefully this helped you understand what B is!)

The equation for the range of a projectile is:
R = [u² sin(2A)]/g, where u is the initial velocity of the projectile and NOT the initial velocity of the marble.

We know that R=x and that sin(2A) = 2sinAcosA = 2cosBsinB = 2(1/3)(2/3)sqrt[2] = (4/9)sqrt[2]. Then:
R = [u² sin(2A)]/g = u²(4/9g)sqrt[2]
(4/3)sqrt[2] r = u²(4/9g)sqrt[2]
r = u²(1/3g) => u²=3rg.
To avoid confusion, I'll replace u with U: U²=3rg.

Now, at the starting position:
KE = 0.5mv² = 0.5mu²
PE = 0

And at the tip of the opening:
KE = 0.5mv² = 0.5mU²
PE = mgh = mg(r+r/3) = (4/3)mrg

Conservation of energy gives:
0.5mu² = 0.5mU² + (4/3)mrg
u² = U² + (8/3)rg
u² = 3rg + (8/3)rg [since]
u² = (17/3)rg
This is the speed the marble should have to DIRECTLY land at the other end of the opening. So any speed greater than this would cause the marble to miss the opening and leave the circle, i.e.
u² > (17/3)rg
as required.