Solutions for Problem Set 11

Last revised 1999/11/02

Chapter 11

Chapter 12

According to the statement of the problem, |p₁| = |p₂|.
Before going through all the cases asked for by the problem, we will get the total angular momentum for the general case of the directions of the two particles not specified and the relative distances from the axis arbitrary. That way we get all the requested cases with one batch of algebra. In particular, x = d/2 on the diagram places the particles equidistant from the axis. Looking at the diagram

r x p₂ = r p sin (theta) 1_z = x p₂ 1_z

where p₂ > 0 if the momentum is upwards on the diagram and p₂ < 0 otherwise, and where 1_z is a unit vector out of the page.

Similarly,

r x p₁ = r p sin (theta) ( - 1_z ) = - (d - x) p₁ 1_z

The total angular momentum is the sum of these two, so

L_tot = [(x - d) p₁ + x p₂ ] 1_z

In parts (a) and (b), the particles travel in opposite directions, so take p₁ = - p₂ = p. Then

L_tot = - d p 1_z which is independent of where the axis is placed, since x does not appear in the answer.

In part (c), take p₁ = + p₂ = p to get

L_tot = (2 x - d) p 1_z

which does depend on where the particles are relative to the axis.

Take the gravitational force to be F = - m g 1_z

Then from projectile motion we have

x = [ v₀ cos (theta₀) ] t

y = [ v₀ sin (theta₀) ] t - (1/2) g t²

v_x = v₀ cos (theta₀ )

v_y = v₀ sin (theta₀ )- g t

L = r x p = m [ x v_y - y v_x ] 1_z

L_z = - (1/2) m v₀ g t² cos (theta₀ ) .

(c) Whereas

r x F = x (-m g) 1_z

torque_z = [ v₀ cos (theta₀) ] t ( - m g )

torque_z = - m v₀ g t cos (theta₀)

Links to image files used in these solutions: 11-73, 73a, 73b, 73c, 73d; 11-76, 76a, 76b, 76c, 76d; 11-85, 85a, 85b, 85c, 85d, 85e; 11-87a, 87b, 87c, 87d, 87e, 87f; 12-15, 15a, 15b, 15c, 15d, 15e, ;15f; (Image 12-15 actually comes from problem 8-39) 12-33; 12-69a, 12-69b, 12-69c,

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