416_PartUniversity Physics Solution

# 416_PartUniversity Physics Solution - 13-8 Chapter 13...

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13-8 Chapter 13 EXECUTE: (a) Initially the energy of the system is 22 2 2 111 1 222 2 (0.175 kg)(0.815 m/s) (155 N/m)(0.0300 m) 0.128 J Em v k x =+ = + = . 2 1 2 kA E = and 2 2(0.128 J) 0.0406 m 4.06 cm 155 N/m E A k == = = . (b) 2 1 max 2 mv E = and max ( 0 . 1 2 8 J ) 1.21 m/s 0.175 kg E v m = . (c) 155 N/m 29.8 rad/s 0.175 kg k m ω = EVALUATE: The amplitude and the maximum speed depend on the total energy of the system but the angular frequency is independent of the amount of energy in the system and just depends on the force constant of the spring and the mass of the object. 13.32. IDENTIFY: 2 1 2 K mv = , grav Um g y = and 2 1 el 2 Uk x = . SET UP: At the lowest point of the motion, the spring is stretched an amount 2 A . EXECUTE: (a) At the top of the motion, the spring is unstretched and so has no potential energy, the cat is not moving and so has no kinetic energy, and the gravitational potential energy relative to the bottom is 2 2 2(4.00 kg)(9.80 m/s )(0.050 m) 3.92 J mgA . This is the total energy, and is the same total for each part. (b) grav spring 0, 0, so 3.92 J UK U = . (c) At equilibrium the spring is stretched half as much as it was for part (a), and so 1 spring 4 (3.92 J) 0.98 J, U 1 grav 2 (3.92 J) 1.96 J, U and so 0.98 J. K = EVALUATE: During the motion, work done by the forces transfers energy among the forms kinetic energy, gravitational potential energy and elastic potential energy. 13.33. IDENTIFY: The location of the equilibrium position, the position where the downward gravity force is balanced by the upward spring force, changes when the mass of the suspended object changes. SET UP: At the equilibrium position, the spring is stretched a distance d . The amplitude is the maximum distance of the object from the equilibrium position. EXECUTE: (a) The force of the glue on the lower ball is the upward force that accelerates that ball upward. The upward acceleration of the two balls is greatest when they have the greatest downward displacement, so this is when the force of the glue must be greatest. (b) With both balls, the distance 1 d that the spring is stretched at equilibrium is given by 1 (1.50 kg 2.00 kg) kd g and 1 20.8 cm d = . At the lowest point the spring is stretched 20.8 cm 15.0 cm 35.8 cm + = . After the 1.50 kg ball falls off the distance 2 d that the spring is stretched at equilibrium is given by 2 (2.00 kg) kd g = and 2 11.9 cm d = . The new amplitude is 35.8 cm 11.9 cm 23.9 cm −= . The new frequency is 1 1 165 N/m 1.45 Hz 2 2 2.00 kg k f m ππ = . EVALUATE: The potential energy stored in the spring doesn&t change when the lower ball comes loose. 13.34. IDENTIFY: The torsion constant κ is defined by z τ κθ = − . 1 2 f I π = and 1/ Tf = . ( ) cos( ) tt θ φ + . SET UP: For the disk, 2 1 2 IM R = . z FR =− . At 0 t = , 3.34 0.0583 rad =Θ= = ° , so 0 = . EXECUTE: (a) (4.23 N)(0.120 m) 8.71 N m/rad 0.0583 rad 0.0583 rad z FR = (b) 1 1 2 1 2(8.71 N m/rad) 2.71 Hz 2 2 2 (6.50 kg)(0.120 m) f R κκ = = . 1/ 0.461 s = = . (c) 2 13.6 rad/s f . ( ) (3.34 )cos([13.6 rad/s] ) = ° . EVALUATE: The frequency and period are independent of the initial angular displacement, so long as this displacement is small.

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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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416_PartUniversity Physics Solution - 13-8 Chapter 13...

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