Linear momentum in a pipe with a 90 degree curve

[Billy Berakus]

Hi,

This is my first thread related to an idea I have been working on and I would really appreciate some help. My education goes no further than high school, so I am doing my best.

I apologize if I am doing this completely incorrectly. I am truly trying to do my best.

To understand direction in this 2D model, I will use compass directions of N, E, S, W.

We have a 1 kg steel ball that is inside a 10 kg pipe in space. The 1 kg mass is traveling at 10 m per s (ms) in a straight line going East inside the pipe that is at rest in comparison to the 1 kg mass. There is no resitance inside the pipe between the steel ball and the pipe.

The pipe has a 90 degree turn in it, in the shape of a 1/4 circle that causes the 1 kg mass to go from an East direction to a South direction. To describe the corner it has an inside diameter of 1 m with the circumference of the 1/4 circle being equal to 1.57 m. (measured from center of pipe)

When the 1 kg mass starts the corner it is going 10 ms but when it finishes it turn of 1.57 m it will be going 6.9 ms, therefor looses 31% of it's momentum due to the corner.

I need to understand how much momentum is transferred to the pipe, and in which direction??

The way I approached the concept is...

We have two forces that will push the pipe both North as well as East. As the 1 kg ball goes South it will push the pipe in equivalence North, however when it goes East it will push East.

I have been doing some experimenting on my air hockey board to make these assumptions.

1) Ep = East Momentum of 1 kg
Ep = mv
Ep = 1 kg x 10 ms
Ep = 10 p

2) Sp = South Momentum of 1 kg (required to push the 1 kg mass south 1 m)
Sp = mv

a) Need to determine velocity requried
t = d/v
0.157 s = 1.57 m / 10 ms (starting velocity 1 kg)
0.228 s = 1.57 m / 6.9 ms (end of 1/4 circle 1 kg)
------
0.192 s Total of both times divided by 2

v = d/t
v = 1 m / 0.192 s (1 m based on radius of 1/4 circle turn)
v = 5.2 ms

b) Sp = mv
Sp = 1 kg x 5.2 ms
Sp = 5.2 p

3) 1 kg Ball 31 % loss on 90 curve = Momentum (Ep & Sp) x 31%

Ep = 10 p x 31%
Ep = 3.1 p

Sp = 5.2 p x 31%
Sp = 1.61 p

Conclusion:

The momentum Sp will be Np on the pipe, so Np + Ep will equal the direction of the pipe.

Am I close?

Thanks!

 

[LightHero]

Yes, you are on the right track. Your conclusion is correct: the momentum lost by the ball (both Ep and Sp) will be transferred to the pipe, resulting in a Np + Ep direction. However, your calculations are slightly off. The velocity of the ball at the end of the curve should be 8.6 ms (not 6.9 ms) and thus the momentum Sp should be 4.3 p (not 5.2 p). The 31% loss would then result in Ep = 2.7 p and Sp = 1.3 p. Therefore, the direction of the pipe should be Np + 2.7p. Hope this helps!