Physics Solution Manual for 1100 and 2101

In other words it remains constant we will use the

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Unformatted text preview: there are no external torques acting on the system, angular momentum will be conserved. I0 ω0 = If ωf 2 2 where I0 = mr0 , and If = mrf . Making these substitutions yields r02 ω0 = rf2 ωf (2) Solving Equation (2) for ωf and substituting into Equation (1) yields: µs g = 4 2 r0 rf ω0 4 rf Solving for rf gives 1/3 ω2 r4 rf = 0 0 µg s 1/3 (2.2 rad/s) 2 (0.30 m)4 = 2 (0.75)(9.80 m/s ) = 0.17 m Chapter 9 Problems 485 69. REASONING To calculate the torques, we need to determine the lever arms for each of the forces. These lever arms are shown in the following drawings: Axis Axis 32.0° 2.50 m 32.0° 2.50 m lT = (2.50 m) cos 32.0° T W lW = (2.50 m) sin 32.0° SOLUTION a. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is Magnitude of torque = W l W = (10 200 N )( 2.5 m ) sin 32° = 13 500 N ⋅ m b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is Magnitude of torque = T l T = ( 62 300 N )( 2.5 m ) cos 32° = 132 000 N ⋅ m 70. REASONING The supernova explosion proceeds entirely under the influence of internal forces and torques. External forces and torques play no role. Therefore, the star’s angular momentum is conserved during the supernova and its subsequent transformation from a solid sphere into an expanding spherical shell: Lf = L0 . The star’s initial and final angular momenta are given by L = I ω (Equation 9.10), where I is the star’s moment of inertia, and ω is its angular velocity. Initially, the star is a uniform solid sphere with a moment of inertia 2 given by I = 5 MR 2 (see Table 9.1 in the text). Following the supernova, the moment of inertia of the expanding spherical shell is I = 2 MR 2 (see Table 9.1 in the text). We will use 3 the angular-momentum-conservation principle to calculate the final angular velocity ωf of the expanding supernova shell. SOLUTION Applying the angular-momentum-conservation principle yields If ωf = I0ω0 {{ Lf L0 or 2 3 M Rf2ωf = 2 5 M 2 R0 ω0 or ωf = 2 3R0 ω0 5Rf2 Substituting R0 = R, Rf = 4.0R, and ω0 = 2.0 rev/d, we obtain the final angular velocity of the expanding shell: 486 ROTATIONAL DYNAMICS ωf = 3 R 2 ω0 ( 5 4.0 R ) 2 = 3ω0 80 = 3 ( 2.0 rev/d ) = 0.075 rev/d 80 71. REASONING According to Newton’s second law for rotational motion, Στ = Iα, the angular acceleration α of the blades is equal to the net torque Στ applied to the blades divided by their total moment of inertia I, both of which are known. SOLUTION The angular acceleration of the fan blades is α= Στ 1.8 N ⋅ m = = 8.2 rad/s 2 2 I 0.22 kg ⋅ m 72. REASONING The drawing shows the forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person’s toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. Likewise, we know that the sum of the torques must be zero. (9.7) FHANDS FFEET Axis W lW = 0.840 m lHANDS = 1.250 m SOLUTION Taking upward to be the positive direction, we have FFEET + FHANDS − W = 0 Remembering that counterclockwise torques are positive and using the axis and the lever arms shown in the drawing, we find Wl W − FHANDS l HANDS = 0 FHANDS = Wl W l HANDS = 584 b N g0.840 m g 392 N b = 1.250 m Substituting this value into the balance-of-forces equation, we find FFEET = W − FHANDS = 584 N − 392 N = 192 N Chapter 9 Problems 487 The force on each hand is half the value calculated above, or 196 N . Likewise, the force on each foot is half the value calculated above, or 96 N . 73. REASONING a. The angular acceleration α is defined as the change, ω − ω0, in the angular velocity divided by the elapsed time t (see Equation 8.4). Since all these variables are known, we can determine the angular acceleration directly from this definition. b. The magnitude τ of the torque is defined by Equation 9.1 as the product of the magnitude F of the force and the lever arm l . The lever arm is the radius of the cylinder, which is known. Since there is only one torque acting on the cylinder, the magnitude of the force can be obtained by using Newton’s second law for rotational motion, Στ = F l = I α . SOLUTION a. From Equation 8.4 we have that α = (ω – ω0)/t. We are given that ω0 = 76.0 rad/s, ω = 1 ω0 = 38.0 rad/s , and t = 6.40 s, so 2 α= ω − ω0 = t 38.0 rad/s − 76.0 rad/s = −5.94 rad/s 2 6.40 s The magnitude of the angular acceleration is 5.94 rad/s2 . b. Using Newton’s second law for rotational motion, we have that Στ = F l = I α . Thus, the magnitude of the force is ( )( ) 0.615 kg ⋅ m 2 5.94 rad/s 2 Iα F= = = 44.0 N l 0.0830 m 488 ROTATIONAL DYNAMICS 0.61 m 74. REASONING The arm, being stationary, is in equilibrium, since it has no translational or angular acceleration. Therefore, the net 0.28 m external force and the net external torque 98 N acting on the arm are zero. Using the fact 0.069 m that the net external torque is zero will allow us to determine the magnitude of the force cg Axis M. The drawing at the right shows three 29° forces: M, the 47-N weight of the arm acting 47 N at the arm’s center of gravity (cg), and the 98-N force that acts upward on the right end M of the arm. The 98-N force is applied to the (0.070 m) sin 29° arm by the ring. It is the reaction force that arises in response to the arm pulling downward on the ring. Its magni...
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