Minimum Velocity needed to clear a hemisphere

In summary, at the point x metres away from the centre of the hemisphere, the height of the slice of hemisphere at that point is sqrt(R^2 - x^2).
  • #1
simpleton
58
0

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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  • #2
Can someone please help me? Thanks
 
  • #3
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
 
  • #4
Thank you very much! :)
 

What is the minimum velocity needed to clear a hemisphere?

The minimum velocity needed to clear a hemisphere is the speed required for an object to travel in a parabolic trajectory and completely exit a hemisphere without hitting the surface. It is affected by factors such as the mass and shape of the object, air resistance, and the radius of the hemisphere.

How is the minimum velocity calculated?

The minimum velocity is calculated using the equation v = √(2gh), where v is the minimum velocity, g is the gravitational acceleration, and h is the height of the hemisphere. This equation assumes no air resistance and a perfectly spherical hemisphere.

Can the minimum velocity be exceeded?

Yes, the minimum velocity is the minimum required to clear the hemisphere, but an object can exceed this velocity and still clear the hemisphere. The actual velocity needed will depend on the specific conditions and variables involved.

What are some real-life applications of minimum velocity needed to clear a hemisphere?

Understanding the minimum velocity needed to clear a hemisphere is important in sports such as javelin throwing, discus throwing, and long jump. It is also relevant in fields such as ballistics and space exploration to determine the necessary trajectory and velocity for an object to reach a desired destination.

How does air resistance affect the minimum velocity needed?

Air resistance can significantly affect the minimum velocity needed to clear a hemisphere. Objects with a larger surface area or less streamlined shape will experience more air resistance, requiring a higher minimum velocity to overcome it. This effect is often taken into account in real-world scenarios.

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