Leaving a circular path

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February 22nd 2008, 02:13 PM
free_to_fly

Leaving a circular path

The question that I'm stuck on is:
A bowl is made from a smooth spherical shell of radius r by cutting away the part which is more that $\ \frac{1}{3}r \$ above the horizontal plane through the centre. A marble is projected from the lowest point of the bowl with speed u. Show that in the subsquent motion the marble will leave the bowl and not fall back into it provided:
$\ u^2 > \frac{{17gr}}{3} \$

I don't know where to being, in fact, I'm not even 100% sure of what the question's saying, so any help would be greatly qppreciated.
February 22nd 2008, 07:12 PM
mr fantastic

Quote:

Originally Posted by free_to_fly

The question that I'm stuck on is:
A bowl is made from a smooth spherical shell of radius r by cutting away the part which is more that $\ \frac{1}{3}r \$ above the horizontal plane through the centre. A marble is projected from the lowest point of the bowl with speed u. Show that in the subsquent motion the marble will leave the bowl and not fall back into it provided:
$\ u^2 > \frac{{17gr}}{3} \$

I don't know where to being, in fact, I'm not even 100% sure of what the question's saying, so any help would be greatly qppreciated.

Quote:

Originally Posted by free_to_fly

A bowl is made from a smooth spherical shell of radius r by cutting away the part which is more that $\ \frac{1}{3}r \$ above the horizontal plane through the centre.

This means that the bowl has a height $h = r + \frac{r}{3} = \frac{4r}{3}$ .
It also means that the bowl curves in on itself so that if the marble leaves the bowl it will perform projectile motion across the top of the bowl. Capisce?

Note that the range of a projectile is given by the formula $\frac{2v^2 \sin \alpha \cos \alpha}{g}$ where v is the initial speed and $\alpha$ is the angle of projection (measured from the horizontal).

You therefore require $\frac{2v^2 \sin \alpha \cos \alpha}{g}$ to be greater than the distance across the bowl (otherwise the marble will land in the bowl. Capisce?).

Let $\theta$ be the angle between the line across the top of the bowl and the radius to the top edge of the bowl (where the marble is going to project from). Then, using simple trigonometry on the obvious 90 degree triangle, you have the following:

1. The distance across the top of the bowl is equal to $2 r \cos \theta$ .

2. $\sin \theta = \frac{r/3}{r} = \frac{1}{3}$ .

3. The angle of projection of the marble (measured from the horizontal line across the top of the bowl is $\alpha = 90 - \theta$ . (This comes from the fact that the marble leaves the bowl at a tangent to the bowl and the fact that the tangent is 90 degrees to the radius).

I'd draw a diagram but it's good for you to draw one and nut this out ...... (that plus the fact that I'm just plain lazy).

To summarise then:

The marble leaves the bowl with initial velocity v and angle of projection $90 - \theta$ to the horizontal. It undergoes projectile motion and you require the range of the marble to be greater than the distance across the bowl.

Mathematically, the requirement is:

$\frac{2v^2 \sin \alpha \cos \alpha}{g} > 2 r \cos \theta$

$\Rightarrow \frac{v^2 \sin (90 - \theta) \cos (90 - \theta)}{g} > r \cos \theta$

$\Rightarrow \frac{v^2 \cos \theta \sin \theta}{g} > r \cos \theta$

$\Rightarrow \frac{v^2 \sin \theta}{g} > r$

$\Rightarrow \frac{v^2}{3g} > r$

$\Rightarrow v^2 > 3gr$ .... (A)

So the value of $v^2$ is needed to finish things off. This can be got via conservation of energy .....

Take G.(ravitational) P.(otential) E.(nergy) of marble at bottom of bowl equal to zero. Then:

(K.E. + G.P.E) of marble at top of bowl = K.E. of marble at bottom of bowl

$\Rightarrow \frac{1}{2} m v^2 + mgh = \frac{1}{2} m u^2$

$\Rightarrow v^2 = u^2 - 2gh = u^2 - 2g\left( \frac{4r}{3} \right) = u^2 - \frac{8gr}{3} = \frac{3u^2 - 8gr}{3}$ .

Substitute this result into (A):

$\frac{3u^2 - 8gr}{3} > 3gr$

$\Rightarrow 3u^2 - 8gr > 9gr$

$\Rightarrow u^2 > \frac{17gr}{3}$ .

Q.E.D. (Cool)
February 23rd 2008, 01:40 AM
free_to_fly

Um where did the formula $ \frac{2v^2 \sin \alpha \cos \alpha}{g} $ come from? I've not come across it before, and I don't think I can quote that in an exam because it's not in the formulae booklet or the textbook.
February 23rd 2008, 02:38 AM
mr fantastic

Quote:

Originally Posted by free_to_fly

Um where did the formula $ \frac{2v^2 \sin \alpha \cos \alpha}{g} $ come from? I've not come across it before, and I don't think I can quote that in an exam because it's not in the formulae booklet or the textbook.

I will derive it for you:

Let the projectile be projected with initial speed v and at an angle $\alpha$ , let it land at the same height from which it was projected and let the only significant force acting on it be gravity. Then the path is symmetrical:

1. Time of flight is equal to twice the time it takes to reach maximum height.

2. Range is equal to twice the horizontal distance travelled when at maximum height.

Vertical component of motion:

$a_y = -g \Rightarrow \frac{dv_y}{dt} = -g \Rightarrow v_y = v \sin (\alpha) - gt$ .

$v_y = v \sin (\alpha) - gt \Rightarrow \frac{dy}{dt} = v \sin (\alpha) - gt \Rightarrow y = v \sin (\alpha) t - \frac{gt^2}{2}$ .

At maximum height, $v_y = 0 \Rightarrow v \sin (\alpha) - gt = 0 \Rightarrow t = \frac{v \sin (\alpha)}{g}$ .

Horizontal component of motion:

$a_x = 0 \Rightarrow \frac{dv_x}{dt} = 0 \Rightarrow v_x = v \cos (\alpha)$ .

$v_x = v \cos (\alpha) \Rightarrow \frac{dx}{dt} = v \cos (\alpha) \Rightarrow x = v \cos (\alpha) t$ .

At maximum height, $t = \frac{v \sin (\alpha)}{g}$ and so $x_{\text{max height}} = v \cos (\alpha) \, \frac{v \sin (\alpha)}{g} = \frac{v^2 \cos (\alpha) \sin (\alpha)}{g}$ .

Therefore:

$\text{Range} = 2 x_{\text{max height}} = \frac{2 v^2 \cos (\alpha) \sin (\alpha)}{g}$ .

Note: From the double angle formula, $\text{Range} = \frac{v^2 \sin (2 \alpha)}{g}$ .

Other derivations are possible.