hw7_sol_s11

# hw7_sol_s11 - Phys 2101 Homework 7 Solution Spring `11...

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Phys 2101 Homework 7 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. The initial angular speed is = (280 rev/min)(2 /60) = 29.3 rad/s. (a) Since the rotational inertia is (Table 10-2(a)) 2 2 (32 kg)(1.2 m) 46.1 kg m I , the work done is 2 2 2 4 1 1 0 (46.1kg m )(29.3 rad/s) 1.98 10 J 2 2 W K I       . (b) The average power (in absolute value) is therefore | | | P W t | 19.8 10 1.32 10 W. 3 3 15 2. (a) We use the parallel-axis theorem to find the rotational inertia:   2 2 2 2 2 2 com 1 1 20 kg 0.10 m 20 kg 0.50 m 0.15 kg m . 2 2 I I Mh MR Mh (b) Conservation of energy requires that Mgh I 1 2 2 , where is the angular speed of the cylinder as it passes through the lowest position. Therefore, 2 2 2 2(20 kg)(9.8 m/s )(0.050 m) 11 rad/s. 0.15 kg m Mgh I 3. From Table 10-2, the rotational inertia of the spherical shell is 2 MR 2 /3, so the kinetic energy (after the object has descended distance h ) is . 2 1 2 1 3 2 2 1 2 2 2 . 2 mv I MR K pulley shell sph

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Since it started from rest, then this energy must be equal (in the absence of friction) to the potential energy mgh with which the system started. We substitute v / r for the pulley’s angular speed and v / R for that of the sphere and solve for v . 2 2 1 1 2 2 3 3 2 2 1 ( / ) (2 /3 ) 2(9.8)(0.82) 1.4 m/s. 1 3.0 10 /((0.60)(0.050) ) 2(4.5)/3(0.60) I M r mgh gh v m I mr M m 4. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic energy of the hoop. The initial kinetic energy is K I mv 1 2 2 1 2 2 (Eq. 11-5), where I = mR 2 is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the speed of its center of mass. Equation 11-2 relates the angular speed to the speed of the center of mass: = v / R . Thus,  2 2 2 2 2 2 1 1 140 kg 0.150 m/s 3.15 J 2 2 v K mR mv mv R which implies that the work required is 0 3.15 J 3.15 J W K     . 5. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic energy is K i = 0 and its initial potential energy is U i = Mgh where 6.0sin30 3.0 m h   (we are using the edge of the roof as our reference level for computing U ). Its final kinetic energy (as it leaves the roof) is (Eq. 11-5) K Mv I f
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