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I have a model that represents a bicycle (a wood block with wheels), and I'm balancing the center of gravity so it's the same as a real bike. However, when the center of mass is kept constant, does the weight of it affect the effect torque has on it when it hits a wall?

I'm planning to measure the angle the back wheel bounces up.

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No. When you hit the wall, the bicycle rotates around the front axis. The angular momentum L that you create for an arbitrary number of mass particles is $$L=\Sigma_i(r_i \times m_iv_i) .$$ If you split location r=R+r_i and v=V+v_i with R and V being center of mass location and velocity, respectively, and r_i and v_i deviation from it, then it can be shown that L does not change when the center of mass does not change.

So, the wood block on wheels should work (in theory).

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  • $\begingroup$ Ah, that's reassuring to hear. However, I'm a complete beginner on the subject, so I don't really understand how something that depends on momentum (with a component of mass) isn't affected by this change in momentum by adding weight? $\endgroup$ – DarkLightA Jan 16 '13 at 15:49
  • $\begingroup$ In your question you didn't mention that you want to add weight to the system. However, it does not matter: Adding mass does not change the angular momentum. If you increase m_i, you will decrease sqrt(v_i*v_i) (keep in mind that r_i and v_i are vectors). The angle of the back wheel will be smaller, but angular momentum is unchanged. $\endgroup$ – Aziraphale Jan 17 '13 at 13:06

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