Minimum Velocity needed to clear a hemisphere

AI Thread Summary
To determine the minimum velocity needed for a ball kicked horizontally from the top of a hemisphere to clear it, the conditions of projectile motion must be analyzed. The height of the ball above the ground decreases as it falls, while the height of the hemisphere at a given horizontal distance can be expressed using the equation for a circle. The critical inequality derived is that the velocity squared must be greater than the product of the radius and gravitational acceleration, v^2 > R*g, to ensure the ball does not contact the hemisphere. This means that if the velocity is less than the calculated minimum, the ball will roll back down the hemisphere. The discussion concludes with the affirmation that this condition must hold true for all time intervals during the ball's trajectory.
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Homework Statement



There is a hemisphere of radius R metres on the ground. You are standing at the very top of the hemisphere and you kick a ball out such that it travels outwards horizontally with speed v metres per second. During the trajectory of the ball, it does not come into contact with the hemisphere at all.
a) What is the minimum v that can fulfil the above conditions?
b) How far away from the centre of the hemisphere will the ball land?

Homework Equations


Projectile Motion Equations


The Attempt at a Solution



Looking at the y-axis : vt = L, where L is the horizontal distance traveled after time t
Looking at the y-axis : H = 0.5*g*t^2, where H is the vertical distance from the top of the hemisphere after time t

Therefore, at time t, the distance of the ball above the ground is R - 0.5*g*t^2.

At the point that is x metres away from the centre, the height of the slice of hemisphere at that point is sqrt(R^2 - x^2).

I can substitute x with vt and get sqrt(R^2 - v^2*t^2)

So now, in order to fulfil the conditions:

R - 0.5*g*t^2 > sqrt(R^2 - v^2*t^2) for t = 0 to the time the ball reaches the ground

Squaring both sides, I get:

R^2 - R*g^2*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2 (I think I don't have to change the bigger-than sign to smaller-than because I know that for this range of values, both sides>0)

So if I make v the subject:

v^2*t^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4
v^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4/t^2
v > sqrt(R^2 + R*g^2*t^2 - 0.25*g^2*t^4)/t

But I still have the t term on the Right Hand Side. How do I get rid of it, or is there another way to do this question?

Thanks all in advance.
 
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Can someone please help me? Thanks
 
simpleton said:

The Attempt at a Solution






So now, in order to fulfil the conditions:

R - 0.5*g*t^2 > sqrt(R^2 - v^2*t^2) for t = 0 to the time the ball reaches the ground

Squaring both sides, I get:

R^2 - R*g^2*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2


there is a mistake: g is not squared:

R^2 - R*g*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2

You can subtract R^2 from both sides:

- R*g*t^2 + 0.25*g^2*t^4 > - v^2*t^2


simpleton said:
So if I make v the subject:

v^2*t^2 > + R*g*t^2 - 0.25*g^2*t^4
v^2 > R*g - 0.25*g^2*t^2


But I still have the t term on the Right Hand Side. How do I get rid of it, or is there another way to do this question?

The inequality has to hold for all t, even at t=0. So v^2>R*g. If v is less, the ball rolls downhill.

ehild
 
Thank you very much! :)
 
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