Physics Solution Manual for 1100 and 2101

The bottom end serves as the axis of rotation the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 80 m/s2 ) ( 0.49 m ) cos 49o = 430 N ( 0.90 m ) sin 36o T 61° L 25° 36° lT Lever arm of the tension force 29. SSM WWW REASONING AND SOLUTION Consider the left board, which has a length L and a weight of (356 N)/2 = 178 N. Let Fv be the upward normal force exerted by the ground on the board. This force balances the weight, so Fv = 178 N. Let fs be the force of static friction, which acts horizontally on the end of the board in contact with the ground. fs points to the right. Since the board is in equilibrium, the net torque acting on the board through any axis must be zero. Measuring the torques with respect to an axis through the apex of the triangle formed by the boards, we have L + (178 N)(sin 30.0°) + fs(L cos 30.0°) – Fv (L sin 30.0°) = 0 2 or 44.5 N + fs cos 30.0° – FV sin 30.0° = 0 so that fs = (178 N)(sin 30.0°) – 44.5 N = 51.4 N cos 30.0° Chapter 9 Problems 457 30. REASONING AND SOLUTION The weight W of the left side of the ladder, the normal force FN of the floor on the left leg of the ladder, the tension T in the crossbar, and the reaction force R due to the right-hand side of the ladder, are shown in the following figure. In the vertical direction −W + FN = 0, so that 2 FN = W = mg = (10.0 kg)(9.80 m/s ) = 98.0 N R In the horizontal direction it is clear that R = T. The net torque about the base of the ladder is 75.0º Στ = – T [(1.00 m) sin 75.0o] – W [(2.00 m) cos 75.0o] + R [(4.00 m) sin 75.0o] = 0 W Substituting for W and using R = T, we obtain T= FN T ( 9.80 N ) ( 2.00 m ) cos 75.0° = 17.5 N ( 3.00 m ) sin 75.0° 31. REASONING The net torque Στ acting on the CD is given by Newton’s second law for rotational motion (Equation 9.7) as Στ = Ι α, where I is the moment of inertia of the CD and α is its angular acceleration. The moment of inertia can be obtained directly from Table 9.1, and the angular acceleration can be found from its definition (Equation 8.4) as the change in the CD’s angular velocity divided by the elapsed time. SOLUTION The net torque is Στ = Ι α. Assuming that the CD is a solid disk, its moment of inertia can be found from Table 9.1 as I = 1 MR 2 , where M and R are the mass and radius 2 of the CD. Thus, the net torque is Στ = I α = ( 1 MR2 )α 2 The angular acceleration is given by Equation 8.4 as α = (ω − ω0 ) / t , where ω and ω0 are the final and initial angular velocities, respectively, and t is the elapsed time. Substituting this expression for α into Newton’s second law yields Στ = ( 1 MR 2 )α = ( 1 MR2 ) 2 2 ( )( ω − ω0 t ) 2 21 rad/s − 0 rad/s −4 = 1 17 × 10−3 kg 6.0 × 10−2 m = 8.0 × 10 N ⋅ m 2 0.80 s 458 ROTATIONAL DYNAMICS 32. REASONING AND SOLUTION a. The net torque on the disk about the axle is Στ = F1R − F2R = (0.314 m)(90.0 N − 125 N) = −11 N ⋅ m b. The angular acceleration is given by α = Στ /I. From Table 9.1, the moment of inertia of the disk is 2 2 2 I = (1/2) MR = (1/2)(24.3 kg)(0.314 m) = 1.20 kg⋅m α = (−11 N⋅m)/(1.20 kg⋅m2) = –9.2 rad / s 2 33. REASONING The moment of inertia of the stool is the sum of the individual moments of inertia of its parts. According to Table 9.1, a circular disk of radius R has a moment of inertia of I disk = 1 M disk R 2 with respect to an axis perpendicular to the disk center. Each 2 thin rod is attached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located at a perpendicular distance from the axis that is equal to the radius of the disk. This means that each of the rods has a moment of inertia of Irod = Mrod R2. SOLUTION Remembering that the stool has three legs, we find that the its moment of inertia is 1 I stool = I disk + 3 I rod = 2 M disk R 2 + 3 M rod R 2 = 1 2 1 0 b.2 kg g0.16 mg+ 3b.15 kg g0.16 mg= b b 2 2 0.027 kg ⋅ m 2 34. REASONING According to Table 9.1 in the text, the moment of inertia I of the disk is 1 2 I = MR 2 , where M is the disk’s mass and R is its radius. Thus, we can determine the mass from this expression, provided we can obtain a value for I. To obtain the value for I, we will use Newton’s second law for rotational motion, Στ = I α (Equation 9.7), where Στ is the net torque and α is the angular acceleration. SOLUTION From the moment of inertia of the disk, we have 1 2 I = MR 2 or M= 2I R2 Using Newton’s second law for rotational motion, we find for I that (1) Chapter 9 Problems Στ = I α or I= 459 Στ α Substituting this expression for I into Equation (1) gives M= 2 I 2Στ = R 2 R 2α (2) The net torque Στ is due to the 45-N force alone. According to Equation 9.1, the magnitude of the torque that a force of magnitude F produces is F l , where l is the lever arm of the force. In this case, we have l = R , since the force is applied tangentially to the disk and perpendicular to the radius. Substituting Στ = FR into Equation (2) gives M= 2Στ Rα 2 = 2 FR Rα 2 = 2 ( 45 N ) 2F = = 5.0 kg Rα ( 0.15 m ) 120 rad/s 2 ( ) 35. REASONING Newton’s second law for rotational motion (Equation...
View Full Document

Ask a homework question - tutors are online