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PHYS-207 Honor’s Section LL, HW8 Solutions 11.7. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where h 6.0sin30 3.0 m (we are using the edge of the roof as our reference level for computing U). Its final kinetic energy is (Eq. 11-5) K f 21 Mv 2 21 I 2 . Here we use v to denote the speed of its center of mass and is its angular speed — at the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we can set v = R = v where R = 0.10 m. Using I 1 2 MR 2 (Table 10-2(c)), conservation of energy leads to 1 1 1 1 3 Mgh Mv 2 I 2 MR 2 2 MR 2 2 MR 2 2 . 2 2 2 4 4 The mass M cancels from the equation, and we obtain hb g c 1 4 1 4 gh 9.8 m s2 3.0 m 63 rad s . R 3 010 . m 3 (b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the origin at the position of the center of mass when the ball leaves the track (the “initial” position for this part of the problem) and take +x leftward and +y downward. The result of part (a) implies v0 = R = 6.3 m/s, and we see from the figure that (with these positive direction choices) its components are v0 x v0 cos30 5.4 m s v0 y v0 sin 30 3.1 m s. The projectile motion equations become x v0 x t and y v0 y t 1 2 gt . 2 We first find the time when y = H = 5.0 m from the second equation: t v0 y v02y 2 gH g 0.74s. b gb g Then we substitute this into the x equation and obtain x 5.4 m s 0.74 s 4.0 m. r F is equal to yFz zFy i zFx xFz j xFy yFx k. then (using Eq. 3-30) we find 11.25. If we write r x i yj zk, d i b gd i (a) Plugging in, we find (b) We use Eq. 3-27, ˆ 3.0m 6.0N 4.0m 8.0N kˆ (50 N m) k. | r F | rF sin , where is the angle between r and F . Now r x 2 y 2 5.0 m and F Fx2 Fy2 10 N. Thus, b gb g rF 5.0 m 10 N 50 N m, the same as the magnitude of the vector product calculated in part (a). This implies sin = 1 and = 90°. 11.39. (a) Since = dL/dt, the average torque acting during any interval t is given by d i avg L f Li t , where Li is the initial angular momentum and Lf is the final angular momentum. avg 0.800 kg m2 s 3.00 kg m2 s 1.47 N m , 1.50s or | avg | 1.47 N m . In this case the negative sign indicates that the direction of the torque is opposite the direction of the initial angular momentum, implicitly taken to be positive. (b) The angle turned is 0t t 2 / 2. If the angular acceleration is uniform, then so is the torque and = /I. Furthermore, 0 = Li/I, and we obtain 2 Li t t 2 / 2 3.00 kg m s 1.50s 1.467 N m 1.50s / 2 I 0.140 kg m 2 2 20.4 rad. (c) The work done on the wheel is W 1.47 N m 20.4rad 29.9 J where more precise values are used in the calculation than what is shown here. An equally good method for finding W is Eq. 10-52, which, if desired, can be rewritten as W L f Li 2 2 2I . (d) The average power is the work done by the flywheel (the negative of the work done on the flywheel) divided by the time interval: Pavg W 29.8 J 19.9 W . t 1.50s 11.58. The initial rotational inertia of the system is Ii = Idisk + Istudent, where Idisk = 300 kg m2 (which, incidentally, does agree with Table 10-2(c)) and Istudent = mR2 where m 60 kg and R = 2.0 m. The rotational inertia when the student reaches r = 0.5 m is If = Idisk + mr2. Angular momentum conservation leads to I disk mR 2 I i i I f f f i I disk mr 2 which yields, for i = 1.5 rad/s, a final angular velocity of f = 2.6 rad/s. 12.5. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a “kink” similar to that shown for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2T sin = F, where is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). In this problem, we have 0.35m 11.5. 1.72 m tan 1 Therefore, T = F/(2sin ) = 7.92 × 103 N. 12.24. As shown in the free-body diagram, the forces on the climber consist of T from the rope, normal force FN on her feet, upward static frictional force f s , and downward gravitational force mg . Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton’s second law to the vertical and horizontal directions, we have 0 Fnet, x FN T sin 0 Fnet, y T cos f s mg . In addition, the net torque about O (contact point between her feet and the wall) must also vanish: 0 net mgL sin TL sin(180 ) O From the torque equation, we obtain T mg sin / sin(180 ). Substituting the expression into the force equations, and noting that f s s FN , we find the coefficient of static friction to be s With f s mg T cos mg mg sin cos / sin(180 ) FN T sin mg sin sin / sin(180 ) 1 sin cos / sin(180 ) . sin sin / sin(180 ) 40 and 30 , the result is 1 sin cos / sin(180 ) 1 sin 40 cos30 / sin(180 40 30) sin sin / sin(180 ) sin 40 sin 30 / sin(180 40 30) 1.19. s 12. 35. We examine the box when it is about to tip. Since it will rotate about the lower right edge, that is where the normal force of the floor is exerted. This force is labeled FN on the diagram that follows. The force of friction is denoted by f, the applied force by F, and the force of gravity by W. Note that the force of gravity is applied at the center of the box. When the minimum force is applied the box does not accelerate, so the sum of the horizontal force components vanishes: F – f = 0, the sum of the vertical force components vanishes: FN W 0 , and the sum of the torques vanishes: FL – WL/2 = 0. Here L is the length of a side of the box and the origin was chosen to be at the lower right edge. (a) From the torque equation, we find F W 890 N 445 N. 2 2 (b) The coefficient of static friction must be large enough that the box does not slip. The box is on the verge of slipping if s = f/FN. According to the equations of equilibrium FN = W = 890 N, f = F = 445 N, s f 445 N 0.50. FN 890 N (c) The box can be rolled with a smaller applied force if the force points upward as well as to the right. Let be the angle the force makes with the horizontal. The torque equation then becomes FL cos + FL sin – WL/2 = 0, F W . 2(cos sin ) We want cos + sin to have the largest possible value. This occurs if = 45º, a result we can prove by setting the derivative of cos + sin equal to zero and solving for . The minimum force needed is F W 890 N 315 N. 2(cos 45 sin 45) 2(cos 45 sin 45) Note: The applied force as a function of is plotted below. From the figure, we readily see that corresponds to a maximum and 45 to a minimum. 0 12.60. (a) Equation 12-8 leads to T1 sin40º + T2 sin = mg . Also, Eq. 12-7 leads to T1 cos40º T2 cos = 0. Combining these gives the expression T2 mg . cos tan 40 sin To minimize this, we can plot it or set its derivative equal to zero. In either case, we find that it is at its minimum at = 50. (b) At = 50, we find T2 = 0.77mg.