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Physics II Homework I CJ Chapter 13; 4, 11, 13, 23, 55, 67, 71 13.4. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body. Solve: (a) The -versus-t graph has a positive slope of 5 rad/s2 from t 0 s to t 2 s and a negative slope of 5 rad/s2 from t 2 s to t 4 s. (b) The angular velocity is the area under the -versus-t graph: d ( x) dt 0 area under graph. dt 13.11. Visualize: Solve: Torque by a force is defined as Frsin where is measured counterclockwise from the r vector to the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force: (30 N)r1 sin1 (20 N)r2 sin2 (30 N)(0.02 m) sin ( 90) (20 N)(0.02 m) sin (90) ( 0.60 N m) (0.40 N m) 0.20 N m Assess: A negative torque causes a clockwise motion of the pulley. 13.13. Visualize: Solve: The net torque on the spark plug is Fr sin 38 N m F(0.25 m)sin(120) F 175.5 N That is, you must pull with a force of 175.5 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque. 13.23. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque () needed to take the plastic disk from i 0 rad/s to f 1800 rpm (1800)(2)/ 60 rad/s 60 rad/s in tf – ti 4.0 s, we need to determine the angular acceleration () and the disk’s moment of inertia (I) about the axle in its center. The radius of the disk is R 10.0 cm. We have 1 1 MR2 (0.200 kg)(0.10 m)2 1.0 103 kg m2 2 2 i 60 rad/s 0 rad/s f i (t f t i ) f 15 rad/s2 tf ti 4.0 s I Thus, I (1.0 103 kg m2)(15 rad/s2) 0.0471 N m. 13.55. Visualize: Solve: We can divide the face of the square rod into small areas dA dx dy with mass dm. Because the mass of the rod is uniformly distributed, dm is the same fraction of M as dA is of A L2. That is dA dm dx dy dm M 2 L2 M L The moment of inertia is I r 2 dm L/2 M L/2 L/2 2 M L/2 M L/2 L/2 ( x y2 ) dx dy 2 x 2dx dy 2 dx y 2dy 2 L L / 2 L / 2 L L / 2 L L / 2 L / 2 L / 2 M x3 2 L 3 L/2 L/2 L/2 M y3 y 2x L L / 2 3 L / 2 L / 2 L/2 L / 2 1 ML2 6 13.67. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize: Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to m1, m2, and the pulley yields the three equations: T1 w1 m1a1 w2 T2 m2a2 T2 R T1R 0.50 N m I Noting that –a2 a1 a, I 21 mp R2 , and a/R, the above equations simplify to T1 m1g m1a m2g T2 m2a 0.50 N m 1 a 1 0.50 N m 1 T2 T1 mp R2 mpa 2 R R R 2 0.060 m Adding these three equations, 1 (m2 m1 )g a m1 m2 mp 8.333 N 2 (m2 m1 )g 8.333 N (4.0 kg 2.0 kg)(9.8 m/s2 ) 8.333 N a 1.610 m/s2 m1 m2 21 mp 2.0 kg 4.0 kg (2.0 kg/2) We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 1 1 y1 y0 v0 (t1 t0 ) a2 (t1 t0 )2 0 1.0 m 0 (1.610 m/s2 )(t1 0 s)2 2 2 t1 2(1.0 m) 1.11 s (1.610 m/s2 ) 13.71. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize: The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation Kf Ugf Ki Ugi is 1 1 1 1 M (v1 )2cm I cm (1 )2 Mgy1 M (v0 )2cm I cm ( 0 )2 Mgy0 2 2 2 2 ( v )2 1 1 2 1 1 0 J 0 J Mgy1 M (v0 )2cm MR2 ( 0 )2cm 0 J Mgy1 M (v0 )2cm MR 2 0 2cm 2 2 3 2 3 R 5 ( v )2 5 5 (5.0 m/s)2 gy1 (v0 )2cm y1 6 0 cm 2.126 m 6 g 6 9.8 m/s2 The distance traveled along the incline is s y1 2.126 m 4.25 m sin30 0.5 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph.