Hovering machine

Classical Mechanics Level pending

Alice has invented a new way of making an object hovering above the ground. The method requires two machines. On the ground, there is the ground machine that keeps throwing balls of mass m = 1 kg m=1\text{ kg} vertically up in the air with velocity v v every τ = 0.1 \tau=0.1 seconds. This machine also catches the balls coming back to it and re-uses them as necessary. Before the balls reach their highest point, they enter into the hovering machine that catches and throws them back with the same velocity as they came to it with. This machine has a weight of M g = 141.5 N Mg=141.5 \text{ N} .

What should be the value of v v ( ( in m/s ) \text{m/s}) for the machine hovering at a height of h = 10 m ? h=10\text{ m}?

Notice that in each machine the direction of the velocity of the balls changes, but the magnitude of the velocity does not need to change (except if we need to re-supply energy that is lost due to air drag and friction). Therefore this method of flying can be very energy-efficient.


The answer is 15.81.

Laszlo Mihaly by Laszlo Mihaly , 57, USA

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1 solution

Laszlo Mihaly
Aug 22, 2018

The velocity of the balls at elevation h h is v ′ = v 2 − 2 g h v'=\sqrt{v^2-2gh} . The momentum change, including the momentum of the ball that leaves downward with the same velocity, is Δ p = 2 m v ′ \Delta p=2mv' . If the balls arrive in time intervals of τ \tau , the average force is F = Δ p / τ = 2 m v ′ / τ F=\Delta p / \tau=2mv'/\tau . We make this equal to the weight:

M g = 2 m v ′ τ = 2 m v 2 − 2 g h τ Mg=\frac{2mv'}{\tau}=\frac{2m\sqrt{v^2-2gh}}{\tau}

The solution is

v = ( M g τ 2 m ) 2 + 2 g h = 15.81 v=\sqrt{\left(\frac{Mg\tau}{2m}\right)^2+2gh}=15.81 m/s

Notice that the force acting on the "ground machine" is actually the combined weight of the "hovering machine" and the balls in the air. To see this we first calculate the time it takes the balls to go up: t = ( v − v ′ ) / g t=(v-v')/g . The number of balls in the air going up is N = t / τ N=t/\tau . There is an equal number of balls coming down and the total mass is M ′ = 2 m g τ ( v − v ′ ) M'=\frac{2m}{g\tau}(v-v') . We re-arrange and use our first equation to express v ′ v' :

2 m v τ = M ′ g + 2 m v ′ τ = ( M ′ + M ) g \frac{2mv}{\tau}=M'g+ \frac{2mv'}{\tau}= (M'+M)g

The left hand side of the equation is the momentum change of the balls in the ground machine. The force acting on the ground machine is F t o t = 2 m v τ = ( M ′ + M ) g F_{tot}=\frac{2mv}{\tau}= (M'+M)g .

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