431_PartUniversity Physics Solution

431_PartUniversity Physics Solution - Periodic Motion 13-23...

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Periodic Motion 13-23 SET UP: (1 ) () / nn d r dr nr −− + =− , for 1 n . EXECUTE: (a) 7 0 92 1 . r dU R F Α dr r r ⎡⎤ ⎛⎞ = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ (b) Setting the above expression for r F equal to zero, the term in square brackets vanishes, so that 7 0 eq eq 1 , R rr = or 77 0e q , Rr = and eq 0 . rR = (c) 19 0 0 7 ( ) 7.57 10 J. 8 A UR R × (d) The above expression for r F can be expressed as 00 22 0 0 ( / )) r Ar r A Fx R x R RR R R ⎛⎞ ⎛⎞ = + + ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ 0 3 0 7 [(1 9( / )) (1 2( / ))] ( 7 / ) . r AA A R x Rx R x R ≈− = = (e) 12 3 0 11 7 /8 . 3 9 1 0 H z . A fk m Rm ππ == = × EVALUATE: The force constant depends on the parameters A and 0 R in the expression for ( ). Ur The minus sign in the expression in part (d) shows that for small displacements from equilibrium, r F is a restoring force. 13.88. IDENTIFY: Apply cm z z I τ α = and cm x x FM a = to the cylinders. Solve for cm . x a Compatre to Eq.(13.8) to find the angular frequency and period, 2 T πω =. SET UP: Let the origin of coordinate be at the center of the cylinders when they are at their equilibrium position. Figure 13.88a The free-body diagram for the cylinders when they are displaced a distance x to the left is given in Figure 13.88b. EXECUTE: cm z z I = 2 1 2 s fR MR = 1 2 s f MR = But cm R a = so 1 cm 2 s f Ma = Figure 13.88b x x Fm a = cm s f kx Ma −= 1 cm cm 2 M ak xM a 3 cm 2 kx Ma = cm (2 /3 ) kMx a = Eq. (13.8): 2 x ax ω (The minus sign says that x and x a have opposite directions, as our diagram shows.) Our result for cm a is of this form, with 2 2/3 kM = and 2/3 . = Thus 2/ 2 / 2. TM k π 3 EVALUATE: If there were no friction and the cylinder didn&t roll, the period would be . M k The period when there is rolling without slipping is larger than this.
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13-24 Chapter 13 13.89. IDENTIFY: Apply conservation of energy to the motion before and after the collision. Apply conservation of linear momentum to the collision. After the collision the system moves as a simple pendulum. If the maximum angular displacement is small, 1 2 g f L π = . SET UP: In the motion before and after the collision there is energy conversion between gravitational potential energy mgh , where h is the height above the lowest point in the motion, and kinetic energy. EXECUTE: Energy conservation during downward swing: 2 1 20 2 2 mgh mv = and 2 0 2 2(9.8 m/s )(0.100 m) 1.40 m/s vg h == = . Momentum conservation during collision: 22 3 () m m V =+ and 2 23 (2.00 kg)(1.40 m/s) 0.560 m/s 5.00 kg V mm = + . Energy conservation during upward swing: 2 f 1 2 Mgh MV = and 2 2 f 2 (0.560 m/s) /2 0.0160 m 1.60 cm 2(9.80 m/s ) hV g = = . Figure 13.89 shows how the maximum angular displacement is calculated from f h .
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431_PartUniversity Physics Solution - Periodic Motion 13-23...

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