Question

even though the solutions are literally there i am confused and do not know how to do any of these problems

ngular Momentum 2. What is the angular momentum of the moon as it revolves around the Earth? Assume the period of the moon's orbit is 27.3 days. Answer: 2.89 × 1034 kg m2/s The position of a particle of mass m with respect to the origin is given by 22. Use r x p to find an expression for the angular momentum with respect to the origin. Hint: From r, use simple calculus to find v. Then, p is mv.... Answer: L=m(-36t2F--36t2Mǐ A 70.0 kg person is rolling on skates at 2.50 m/s. He grabs a pole and starts to revolve around it at arm's length (1.10 m) at the same speed. Prove that the angular momentum is the same before (L-rxp) and after (L lo). 23. rsine -R) Solution: Recall that although r and θ are changing during the approach, rin() R. So, Calculation from Io is straightforward where I is mR2 and ω is 2.27 rad/s. You can see that both have a counterclockwise sense (out of the page). Answer: 192 kg m/s, out of the page. A bullet of mass m hits a long, thin rod of length L and mass M which is hanging from its end and free to swing upward. The bullet strikes the end of the rod, stops completely upon was the bullet moving? If this causes the rod's center of mass to rise by h, how fast impact, and falls away. Solution: We have a collision (momentum problem) followed by a change in height (energy problem). Conservation of momentum gives L=L' psin 8-lo The kinetic energy it has after the collision becomes potential energy at the top of the swing 2 3 Substitute for o. You get 1 M Atlantic Cape Community College ScienggD

Answer 1

ms mess Moon 84400 000 예) 2. Moment 따 next?a about. o 2 ML ular momentum ?--9.ott tut.夕 (b) 3m From cosseryeetion enengy -36 tim HrLx? 2 3m 3 )リ 3 m L