Problem 4. This problem is about calculating angular momentum in three unrelated

Question

Problem 4. This problem is about calculating angular momentum in three unrelated situations. (i) A particle of mass 4.00 kg is moving in a straight line with a constant speed of 5.00 m/s. The closest distance of the particle from the point P is 1.60 m. (The trajectory as well as the point P are in the plane of the page.) (a) What is the angular momentum (magnitude and direction) of the particle about P when it is at the point of closest approach to P? (b) Show that the angular momentum of the particle when it is at the point making the angle theta1 = 30 degree and at the point making the angle theta2 = 60 degree is the same as in part (a). (ii) Consider the Earth to be a uniform, solid sphere rotating about an axis passing through its center. What is the angular momentum of the Earth about this axis? (The astronomical values you need can be found in Appendix F of the textbook or online.) iii) The lowest energy orbital of an electron in a Hydrogen atom is a circle of radius 5.29 x 10^-11 m. The angular momentum of the electron in that orbital is 1.055 x 10^-34 j dot s. What is the speed (v) of the electron? (Again, you can find a physical quantity you need either in appendix F or online.)

Explanation / Answer

i) L = m*v*r*sin theta

theta = the angle between v and position vector r

a) L = m*V*y*sin90 = 4*5*1.6 = 32 kg m^2/s

b) at theta = 30

L = m*v*r1*sin30 = m*v*y = 32

L = m*v*r2*sin(60) = m*v*y = 32

ii) w = 2*pi/T = (6.28)/(24*60*60) = 7.3e-5

I = (2/5)*M*R^2 = (2/5)*5.98e24*6371000*6371000 =

9.7090421272e+37

L = I*w = 9.7090421272e+37*7.3e-5 = 7.08e33 kg m^s /s

iii) L = m*v*R = 1.055e-34

v = L/mR = (1.055e-34)/(9.1e-31*5.29e-11) = 2.19*10^6 m/s

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