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Spinning spacecraft

The application of conservation of angular momentum crops up in many areas of life ... even NASA use it to send satellites into deep space.

The Mars Pheonix mission shows a really nice example of just that. As the spacecraft leaves Earth orbit en route to Mars, it is spun (a technique known as ’rifling’) to more accurately achieve the desired trajectory.

Once safely en route, the rotation is arrested by weights on cords being fired out from the spacecraft, just like the way that an ice skater pulls out of a spin by spreading his or her arms.

You can see an animation of this at about 1:20 in the video below

   

So, your task is to

(a) Calculate the thrust (force) of the rockets used to rotate the landing craft when preparing for rifling, getting up to 2 revolutions per second in 15s.

(b) Calculate the mass of the ejected weights used to reduce the rotation to 0.2 rps, if the cords which attach them to the craft are 5m long

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Hint

You’ll need some data : the lander has a mass of 350 kg, and measures 2.2 m tall by 5.5 m long with its solar panels deployed.
 

Solution

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Solution

For the first part.....

Our strategy is to calculate the required angular acceleration to achieve the desired rotation and from this, find an expression for the torque that produces this acceleration.

If we assume constant angular acceleration, we can use ω=ω0+αtα=ω/t.

We can then use the rotational equivalent of Newton’s second law, τ=Iα and we know that torque, τ, is r×F.

Thus, $F r \sin \theta = I \alpha \Rightarrow Fr = \frac{I\omega}{t}$

If we assume that the landing craft is a solid cylinder, then it has moment of inertia, I, given by I=MR2/2.

A thick-walled shell is better

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A thick-walled shell is better

It’s slightly more realistic to think of it as a shell with internal and external radii. In this case the expression becomes $I = \frac{1}{2} M (R_1^2 + R_2^2)$
 

$F = \frac{I\omega}{rt} = \frac{mr^2\omega}{2rt} = \frac{m\omega r}{2t}$

Two revolutions per second is 4πrads-1, hence

F160N

Since there are two rockets, that’s 80N each.

For the second part .....

We can say here that angular momentum is going to be conserved, so our strategy is to calculate the new moment of intertia and hence the mass of the two weights.

$$I_1\omega_1 = I_2\omega_2 \Rightarrow I_2 = I_1 \frac{\omega_1}{\omega_2}$$

$$I_2 = \frac{MR^2}{2} \times \frac{2}{0.2} = 5MR^2$$

I2 is the original moment of interia of the craft, plus the ’extrta bit’ from the ejected weights. If we let m be the mass of the ejected braking weights and r be the distance from the rotation axis, then

$$I_2 = \frac{MR^2}{2} + 2mr^2$$

Equating these two expressions for I2 gives

$$5MR^2 = \frac{MR^2}{2} + 2mr^2 \Rightarrow m = \frac{9MR^2}{4r^2} = 26kg$$

(Note that r here = 5 + 1.1 = 6.1 m)