Unbalanced Thrust
A uniform disk sits on a smooth floor with a rocket strapped to its perimeter. One second after the rocket ignites, how far is the disk's center from where it started, in meters?
Details and Assumptions:
- The mass of the disk is M = 1 k g , its radius is r = 0 . 5 m , and the rocket provides a constant thrust of T = 1 0 N after it ignites.
- Neglect the mass and size of the rocket.
- Submit your answer to to 3 decimal places.
The answer is 1.808.
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1 solution
Why does the center move?
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Because a force is acting on the disk. Total force equals rate of change of momentum of the centre of mass (and total torque equals rate of change of angular momentum about the centre of mass). I am applying these two principles.
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I actually disagree. Unless the thrust vector is constant in an inertial frame, the force is purely tangential and would have no component through the disk's CM in a rotating frame. The center of the disk should not move at all.
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@Josh Ross – The force does not have to go through the center and the disk will still move (accelerate). In fact, the magnitude of the acceleration of the CM is entirely independent of the line of force relative to the CM. The line of force is important only for the rotational motion.
Think of the spinning ball that people hit in ping pong or tennis.
@Josh Ross – I actually totally agree with Laszlo...
You could prevent the center from moving by providing a proper "force couple"
I did exactly the same equations of motions, but my numerical integration yielded 1.64m. Since Steve and Mark agrees on the result, and mine is different, I am still looking for an error in mine.
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Are you getting good convergence as the time step decreases?
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Yes it converges well. I am doing a double integration so it may have an accumulation of some other systematic error. I checked the Fresnel integrals with Wolfram Alpha and I got your result. Excellent problem, BTW.
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@Laszlo Mihaly – If I have convinced you about the way to turn the double integral into a single one, try your system on the single integral. Numerical calculations for double/iterated integrals can be more sensitive...
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@Mark Hennings – That is exactly what I am planning to do. And yes, that was a neat trick with the integrals. My double integration is a brute force method, and certainly more prone to errors.
@Laszlo Mihaly – Thanks. I liked your "spinning" problem. A worthy companion to this one
Mark, I do not understand how did you do the last step, ∫ 0 1 d t ∫ 0 t cos ( 2 0 u 2 ) d u = ∫ 0 1 ( 1 − u ) cos ( 2 0 u 2 ) d u . This looks too simple to me. It looks to me that the ∫ . . . . d u yields a Fresnel integral, and you have to do another integral over the Fresnel function.
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Reverse the order of integration...
See h e r e for an understanding that
∫ 0 1 ( ∫ 0 u f ( t , u ) d u ) d t = ∫ 0 1 ( ∫ u 1 f ( t , u ) d t ) d u
Then in our example: ∫ 0 1 ( ∫ 0 u cos ( 2 0 u 2 ) d u ) d t = ∫ 0 1 ( ∫ u 1 cos ( 2 0 u 2 ) d t ) d u = ∫ 0 1 ( 1 − u ) cos ( 2 0 u 2 ) d u
can anyone point out how he went from x'' = -10sin(20t^2) to x=int(dt(int10*(-sin20u^2)du)) in the last equation.
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x ( 1 ) = x ( 0 ) + ∫ 0 1 x ′ ( t ) d t
x ( 0 ) = 0
x ′ ( t ) = x ′ ( 0 ) + ∫ 0 t x ′ ′ ( u ) d u
x ′ ( 0 ) = 0
x ′ ( t ) = ∫ 0 t − 1 0 sin ( 2 0 u 2 ) d u
Plugging into our first equation:
x ( 1 ) = ∫ 0 1 ∫ 0 t − 1 0 sin ( 2 0 u 2 ) d u d t
The dt was moved to the beginning through a notational variant that I find absurd.
Where can I find the equations of motion ?
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The equations shown above are essentially just Newton's Second Law, for translation and rotation
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Can you make explicit the equations applied? I’d be very grateful!
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@Alessandro Mattia – The first, vector, equation in my solution just states that the mass (1) times the acceleration is the force ( of magnitude 10, in a direction tangential to the disc). The second, scalar equation states that the moment of inertia (1/8) times the angular acceleration is equal to the torque (10 x 0.5) of the thrust about the centre of the disc.
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@Mark Hennings – got it! thanks for your patient!
Set up coordinate axes with the origin at the centre of the disk, and the rocket on the positive x -axis initially. If ( x , y ) are the coordinates of the centre of the disk, and if θ is the angle that the radius from the centre of the disk to the rocket makes with the positive x -axis, then we have the equations of motion ( y ¨ x ¨ ) = 1 0 ( cos θ − sin θ ) 8 1 θ ¨ = 5 Since the system starts at rest, we see that θ = 2 0 t 2 , and hence, after 1 second, ( y x ) = ∫ 0 1 d t ∫ 0 t 1 0 ( cos ( 2 0 u 2 ) − sin ( 2 0 u 2 ) ) d u = 1 0 ∫ 0 1 ( 1 − u ) ( cos ( 2 0 u 2 ) − sin ( 2 0 u 2 ) ) d u These integrals can be evaluated in terms of Fresnel integrals, and numerical evaluation gives x 2 + y 2 = 1 . 8 0 7 7 8 .