Answers for Problem Set 11

Chapter 11

#73.

(a)

tau = (2/5) m R² (omega _f / t _f )

tau ₁ = 0.69 N m

tau ₂ = 9.84 N m

(b)

F = (2/5) m R omega _f / t _f

F₁ = 3.05 newton

F₂ = 11.5 newton

#76. [My answer agrees with an older edition of the text]:

(a) alpha = 2 theta / t²

(b) a = 2 R theta / t² [same for each block]

(c)
T₁ = M (g - 2 R theta / t² )

T₂ = M g - (2 theta / t² ) (M R + I / R )

(a) 4.8 x 10⁵ N

(b) 1.1 x 10⁴ N m

(c) 1.3 x 10⁶ Joule

#87 [My answers agree with text]

(a) a_c = 3 g [1 - cos (theta) ]

(b) (3/2) g sin (theta)

(c) theta = 41.8^o = 0.73 radians

Chapter 12

#15. Text's answers:

(a) 2.7 R

(b) (50/7) m g

#33 Proof.

#41 [The text gives absolute magnitudes of dL/dt and torque]

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

(b) dL_z/dt = m g t v₀ cos (theta₀)

(c) torque_z = m g t v₀ cos (theta₀)

#69
cos (theta) = 1 - 6 h m²/ [ l_p (2 m + M) (3 m + M ) ]

I have used l_p for the length of the pendulum to improve the clarity of the printing. This expression has the correct limits for m = 0, M = 0, and h = 0. Note that g has dropped out because there is no way to cancel out its factor of sec ^-2 .