Physics Solution Manual for 1100 and 2101

0 kg 60 ms 2 mr 2 54 j the kinetic energy of the

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Unformatted text preview: that Rod A = 1 2 KE R = 1 2 = Rod B ML ω ch 0 4 b.66 kg g0.75 m gb.2 rad / sg= b I ω = c ML h ω 0 4 b.66 kg g0.75 m gb.2 rad / sg= b 1 KE R = 2 I A ω 2 = 1 6 2 1 2 2 2 2 B 1 2 1 3 2 2 3. 3 J 2 2 2 1.1 J 51. SSM REASONING AND SOLUTION a. The tangential speed of each object is given by Equation 8.9, v T = rω . Therefore, For object 1: vT1 = (2.00 m)(6.00 rad/s) = 12.0 m / s For object 2: vT2 = (1.50 m)(6.00 rad/s) = 9.00 m / s For object 3: vT3 = (3.00 m)(6.00 rad/s) = 18.0 m / s b. The total kinetic energy of this system can be calculated by computing the sum of the kinetic energies of each object in the system. Therefore, Chapter 9 Problems 473 2 2 1 1 1 KE = 2 m1 v12 + 2 m2 v 2 + 2 m3 v 3 KE = 1 2 ( 6.00 kg)(12.0 m / s) 2 + ( 4 .00 kg)(9.00 m / s) 2 + ( 3.00 kg)(18.0 m / s) 2 = 1.08 × 10 3 J c. The total moment of inertia of this system can be calculated by computing the sum of the moments of inertia of each object in the system. Therefore, I = ∑ mr 2 = m1 r12 + m2 r22 + m3 r32 I = ( 6.00 kg)(2.00 m) 2 + ( 4 .00 kg)(1.50 m) 2 + ( 3.00 kg)(3.00 m) 2 = 60.0 kg ⋅ m 2 d. The rotational kinetic energy of the system is, according to Equation 9.9, 1 1 KE R = 2 Iω 2 = 2 (60.0 kg ⋅ m 2 )(6.00 rad / s) 2 = 1.08 × 10 3 J This agrees, as it should, with the result for part (b). 52. REASONING Each blade can be approximated as a thin rod rotating about an axis perpendicular to the rod and passing through one end. The moment of inertia of a blade is given in Table 9.1 as 1 ML2 , where M is the mass of the blade and L is its length. The total 3 moment of inertia I of the two blades is just twice that of a single blade. The rotational kinetic energy KER of the blades is given by Equation 9.9 as KE R = 1 I ω 2 , where ω is the 2 angular speed of the blades. SOLUTION a. The total moment of inertia of the two blades is 2 I = 1 ML2 + 1 ML2 = 3 ML2 = 3 3 2 3 ( 240 kg )( 6.7 m )2 = 7200 kg ⋅ m2 b. The rotational kinetic energy is KE R = 1 I ω 2 = 2 1 2 ( 7200 kg ⋅ m2 ) ( 44 rad/s )2 = 7.0 × 106 J 53. REASONING The rotational kinetic energy of a solid sphere is given by Equation 9.9 as KE R = 1 I ω 2 , where I is its moment of inertia and ω its angular speed. The sphere has 2 translational motion in addition to rotational motion, and its translational kinetic energy is KE T = 1 mv 2 (Equation 6.2), where m is the mass of the sphere and v is the speed of its 2 474 ROTATIONAL DYNAMICS center of mass. The fraction of the sphere’s total kinetic energy that is in the form of rotational kinetic energy is KER/(KER + KET). SOLUTION The moment of inertia of a solid sphere about its center of mass is I = 2 mR 2 , 5 where R is the radius of the sphere (see Table 9.1). The fraction of the sphere’s total kinetic energy that is in the form of rotational kinetic energy is 1 Iω 2 KE R 2 = = KE R + KE T 1 I ω 2 + 1 mv 2 2 2 ( 1 2 mR 2 25 1 2 mR 2 ω 2 25 ( ) )ω2 + 1 mv 2 2 = 2 R 2ω 2 5 2 R 2ω 2 + v 2 5 Since the sphere is rolling without slipping on the surface, the translational speed v of the center of mass is related to the angular speed ω about the center of mass by v = Rω (see Equation 8.12). Substituting v = Rω into the equation above gives KE R KE R + KE T = 2 R 2ω 2 5 2 R 2ω 2 + v 2 5 = 2 5 2 5 R 2ω 2 R 2ω 2 + ( Rω ) 2 = 2 7 54. REASONING Because we are ignoring frictional losses, the total mechanical energy E = 1 mv2 + 1 Iω 2 + mgh of both objects is conserved as they roll down the hill. We will 2 2 apply this conservation principle twice: first, to determine the height h0 − hf of the hill, and second, to determine the translational speed vf of the frozen juice can at the bottom of the hill. Both the basketball and the frozen juice can roll without slipping, so the translational speed v of either one is related to its radius r and angular speed ω by v = rω (Equation 8.12). SOLUTION a. Applying the energy conservation principle to the basketball, we obtain 1 mv 2 + 1 I ω 2 + mgh = 1 mv 2 + 1 I ω 2 + mgh f f f 0 0 0 2 2 2 2 144424443 144424443 Ef (1) E0 The basketball starts from rest, so we have v0 = 0 m/s and ω0 = 0 rad/s. The basketball is a thin-walled spherical shell, so its moment of inertia is given by I = 2 mr 2 (see Table 9.1 in 3 the text). Substituting these values along with ωf = vf r (Equation 8.12) into Equation (1) yields 1 2 m vf2 + 1 2 ( 2 3 ) 2 v m r 2 f + m ghf = 0 + 0 + m gh0 r or 1 v2 2f + 1 vf2 = 5 vf2 = g ( h0 − hf ) 3 6 Chapter 9 Problems 475 Solving for the height h0 − hf of the hill, we obtain h0 − hf = 5vf2 6g = 5 ( 6.6 m/s ) ( 2 6 9.80 m/s 2 ) = 3.7 m b. In Equation (1), we again substitute v0 = 0 m/s and ω0 = 0 rad/s, but this time we use I = 1 mr 2 for the moment of inertia (see Table 9.1 in the text), because the frozen juice can 2 is a solid cylinder: 1 2 m vf2 + 1 2 ( 1 2 ) 2 v m r 2 f + m ghf = 0 + 0 + m gh0 r or 1 v2 2f + 1 vf2 = 3 vf2 = g ( h0 − hf ) 4 4 The final translational speed of the frozen juice can is vf2 = 4 g ( h0 − hf ) 3 or vf = 4 g ( h0 − hf ) 3 = ( ) 4 9.80 m/s 2 ( 3.7 m ) 3 =...
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