fisica1-youn-e-freedman-exercicios-resolvidos

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10.43: a) From conservation of angular momentum, 2 I1 ( 1 2) MR ω2 = ω1 = ω 2 1 2 I + mR 1 2 MR + mR 0 ( 70) ( ) 3.0 rad s = = 1.385 rad s 1+ 2 120 2 1 = ω1 1+ 2m M or 1.39 rad s to three figures 2 2 1 = = 2 = ( 0 ) ω2 = J. K and b) ( 1 2)( 1 2)( 120 kg)( 2.00 m) ( 3.00 rad s) 1.80 kJ, 2 2 ( 1 2) I + ( 70 kg)( 2.00 m) 499 K In changing the parachutist’s horizontal component of velocity and slowing down the turntable, friction does negative work. 10.44: Let the width of the door be l; ω = = mv( l 2) = 2 2 ( 1 3) Ml + m( l 2) ( 0.500 kg)( 12.0 m s)( 0.500 m) 2 ( 1 3)( 40.0 kg)( 1.00 m) + ( 0.500 kg)( 0.500 m) L I 2 = 0.223 rad s. Ignoring the mass of the mud in the denominator of the above expression gives ω = 0.225 rad s, so the mass of the mud in the moment of inertia does affect the third significant figure. 10.45: Apply conservation of angular momentum L r , with the axis at the nail. Let object A be the bug and object B be the bar. Initially, all objects are at rest and L = 0 1 . Just after the bug jumps, it has angular momentum in one direction of rotation and the bar is rotating with angular velocity ω in the opposite direction. L 2 1 = m L = L 2 3mAv ωB = m r v r − I ω A A gives m v B A B A B A B where r = 1.00 m and I r = 1 3 m B = 0.120 rad s r 2 ω B B = 1 3 m B r 2
10.46: (a) Conservation of angular momentum: m v d = −m vd + 1 0 1 2 m 3 2L ω 1 ⎛ (3.00 kg)(10.0 m s)(1.50 m) = −(3.00 kg)(6.00 m s)(1.50 m) + ⎜ 3 ⎝ ω = 5. 88 rad s 1 90.0 N 9.80 m s 2 ⎞ 2 ⎟ (2.00 m) ω ⎠ (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. 10.47:
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Física I - Sears, Zemansky, Young
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5 1.1: 1mi × ( 5280ft mi) × ( 12
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1.15: a) If a meter stick can measu
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8 11 1.28: The moon is about 4× 10
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1.34: r o o 1.35: A ; A x = ( 12.0
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1.39: Using components as a check f
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1.43: a) The magnitude of A r + B r
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1.48: a) ˆ ˆ ˆ 2 2 2 i + j + k =
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1.53: Use of the right-hand rule to
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1.59: a) To three significant figur
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1.65: Let D r be the fourth force.
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1.68:a) R = A + B + C x R y = x x x
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1.70: The third leg must have taken
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−1 200−20 o 1.73: a) Angle of f
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1.75: Let +x be east and +y be nort
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1.78: (a) Take the beginning of the
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1.83: a) Parallelog ram area = 2 ×
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1.88: a) This is a statement of the
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1.91: A r and C r are perpendicular
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1.96: a) b) i) In AU, ii) In AU, ii
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Capítulo 2
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2.6: The s-waves travel slower, so
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2.14: (a) The displacement vector i
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2.18: a) The velocity at t = 0 is a
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2.20: a) The bumper’s velocity an
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2.26: a) x 0 < 0, v 0x < 0, a x < 0
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2.28: a) Interpolating from the gra
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2.31: a) At t = 3 s the graph is ho
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2.35: a) b) From the graph (Fig. (2
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2.37: a) The car and the motorcycle
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2.43: a) Using the method of Exampl
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2.45: a) v y = v0 y − gt = ( −6
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2.49: a) For a given initial upward
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2.51: a) From Eqs. (2.17) and (2.18
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2.53: a) The change in speed is the
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.0 m 2.56: a) 1.25m s. 25 = 20.0 s
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1 8 m ⎛ 365 d ⎞⎛ ⎞⎛ 2.62:
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2.66: a) The simplest way to do thi
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2 2.69: a) x = 1 2) a t , and with
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2.71: a) Travelling at 20 m s , Jua
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2.77: Let t 1 be the fall for the w
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2.82: a) 2.83: a) From Eq. (2.14),
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2 2.86: a) The helicopter accelerat
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2.90: a) Suppose that Superman fall
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2 2.93: The velocities are vA = α
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1 2.96: The time spent above y max
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2.98: a) Let the height be h and de
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3.1: a) v (5.3 m) − (1.1m) = 1.4
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3.7: a) b) r ˆ ˆ 2 v = αi − 2
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3.11: Take + y to be upward. Use Ch
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3.15: a) Solving Eq. (3.17) for v =
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v y . g 2 9.80 m s 0 16.0 m s 3.18:
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3.22: Substituting for t in terms o
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3.25: Take + y to be downward. a) U
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3.34: a) a 2 rad = ( 3 m/s) /(14 m)
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3.45: a) The a = 0 and a y = −2β
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3.48: a) The equations of motions a
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3.52: a) Setting y = −h in Eq. (3
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3.58: Equation 3.27 relates the ver
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3.60: a) This may be done by a dire
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3.64: Combining equations 3.25, 3.2
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3.66: (a) v x ( runner) = vx (ball)
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3.69: Take + y to be upward. a) The
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3.73: a) The acceleration is given
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3.76: a) dv dt = = d dt (1/ 2) vxax
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3.85: The three relative velocities
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3.90: As in the previous problem, t
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3.92: The x-position of the plane i
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4.1: a) For the magnitude of the su
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2 4.11: a) During the first 2.00 s,
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4.24: (a) Each crate can be conside
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4.27: a) b) For the chair, a y = 0
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4.33: a) The resultant must have no
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2 4.39: a) Both crates moves togeth
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4.43: a) The engine is pulling four
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4.48: a) The net force on a point o
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4.51: a) L is the lift force b) ∑
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2 4.54: The velocity as a function
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4.57: In this situation, the x-comp
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5.1: a) The tension in the rope mus
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5.14: a) In level flight, the thrus
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2L 5.20: Similar to Exercise 5.16,
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5.28: a) The stopping distance is 2
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5.35: First, determine the accelera
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5.40: Differentiating Eq. (5.10) wi
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5.49: a) Setting arad = g in Eq. (5
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5.56: The tension in the lower chai
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5.63: (Denote F r by F.) a) The for
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5.67: Consider the forces on the pe
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5.70: It’s interesting to look at
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5.72: The key idea in solving this
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5.77: See Exercise 5.40. a) The max
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5.82: We take the upward direction
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5.85: Denote the magnitude of the a
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5.92: (a) There is a contact force
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5.96: (a) 80 mi h is 35 .7 m s in S
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5.101: a) The rock is released from
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mg mg 5.103: Without buoyancy, kv t
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5.106: (a) To find find the maximum
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5.109: a) For the same rotation rat
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5.116: a) Differentiating twice, a
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5.120: a) There are many ways to do
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5.124: For convenience, take the po
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5.126: In all cases, the tension in
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6.1: a) ( 2.40 N) (1.5 m) = 3.60 J
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6.10: a) From Eq. (6.6), 1 2 5 ⎛
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6.18: As the example explains, the
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6.27: a) The friction force is µ k
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6.35: a) The static friction force
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6.42: The initial and final kinetic
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6.53: a) In terms of the accelerati
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6.58: The work per unit mass is ( W
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6.66: Let x be the distance past P.
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6.71: The velocity and acceleration
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6.79: In terms of the bumper compre
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6.85: f = .25 mg soW f = W = −(0.
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6.94: a) The number of cars is the
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(8.00 hp)(746 W/hp) 6.99: a) = 358
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6.101: a) Denote the position of a
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6.104: From F = m a, Fx = max, Fy =
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7.1: From Eq. (7.2), mgy = (800 kg)
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7.10: (a) The flea leaves the groun
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2 7.16: U = 1 ky , where y is the v
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1 2 1 2 7.24: From kx = mv , the re
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dU 3 3 7.34: From Eq. (7.15), F =
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7.39: a) At constant speed, the upw
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1 2 7.46: a) U − U = mg( h − 2R
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7.49: a) K 1 + U1 + Wother = K2 + U
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7.54: To be at equilibrium at the b
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7.57: The two design conditions are
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7.62: a) The skier’s kinetic ener
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1 2 1 2 7.69: With U = 0 , K = 0 K
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7.74: a) From either energy or forc
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2 2 2 2 7.78: a) a = d x dt = −ω
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3 7.83: a) Along this line, x = y ,
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7.86: a) The slope of the U vs. x c
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Capítulo 8
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8.7: ∆p ∆t ( 0.0450 kg)( 25.0 m
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8.14: The impluse imparted to the p
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8.21: a) Taking v A and v B to be m
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8.25: m + m )(3.00 m s) = m (4.00 m
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8.31: Use conservation of the horiz
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8.37: Let +y be north and +x be sou
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8.43: a) In Eq. (8.24), let mA = m
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8.52: It turns out to be more conve
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8.63: The total momentum must be ze
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8.69: (This problem involves solvin
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8.72: a) The stuntman’s speed bef
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8.76: Just after the collision: ∑
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8.80: a) From the derivation in Sec
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8.83: a) In terms of the primed coo
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8.86: (a) For momentum to be conser
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8.90: a) For the x - and y - direct
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8.96: The trick here is to realize
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8.98: The two fragments are 3.00 kg
Page 311 and 312: 8.101: a) If the objects stick toge
Page 313 and 314: 8.107: a) For t < 0 the rocket is a
Page 315 and 316: 8.111: a) The tension in the rope a
Page 317 and 318: 9.1: a) 1.50 m = 0.60 rad = 34.4°
Page 319 and 320: 2 9.10: a) ω = ω + α t = 1.50 ra
Page 321 and 322: 9.18: The following table gives the
Page 323 and 324: 9.24: From a rad = ω 2 r, 4 which
Page 325 and 326: vr 5.00 m s 9.33: The angular veloc
Page 327 and 328: 9.40: a) In the expression of Eq. (
Page 329 and 330: 2π 2 2 9.49: a) ω = T , so Eq. (9
Page 331 and 332: 9.60: For this case, dm = γ dx. a)
Page 333 and 334: dθ 2 2 3 2 9.65: a) ω z = = 2γt
Page 335 and 336: 9.70: a) The angular acceleration w
Page 337 and 338: 9.75: I approximate my body as a ve
Page 339 and 340: 9.81: a) Consider a small strip of
Page 341 and 342: 9.84: Taking the zero of gravitatio
Page 343 and 344: 9.90: a) In the case that no energy
Page 345 and 346: 9.94: a) From the parallel-axis the
Page 347 and 348: 9.99: a) Following the hint, the mo
Page 349 and 350: Capítulo 10
Page 351 and 352: 10.6: (a) τ A = (50 N)(sin 60° )(
Page 353 and 354: 1 2 2 = m 2 8.25 kg 0.0750 m = 0.02
Page 355 and 356: 10.23: n = mg cos α mg sin θ −
Page 357 and 358: 10.26: a) The angular speed of the
Page 359 and 360: 10.32: I = mL 1 2 1 2 2 = ( 117 kg)
Page 361: 10.40: The skater’s initial momen
Page 365 and 366: 2 2π rad 10.51: a) Solving Eq. (10
Page 367 and 368: 2 2θ 2θ 2θ I 10.57: t = = = . α
Page 369 and 370: 10.63: The net torque on the pulley
Page 371 and 372: 2 10.67: For the disk, K = (3 4) Mv
Page 373 and 374: 10.72: (a) Στ = Iα and a = Rα T
Page 375 and 376: 10.75: a) Use conservation of energ
Page 377 and 378: 10.79: a) v = 10K 7m = ( 10)( 0.800
Page 379 and 380: 10.84: (a) Στ = Iα 1 1 2 mg cos
Page 381 and 382: 10.90: Angular momentum is conserve
Page 383 and 384: 10.95: a) The initial angular momen
Page 385 and 386: 10.100: a) The distance from the ce
Page 387 and 388: 10.102: Denoting the upward forces
Page 389 and 390: 11.1: Take the origin to be at the
Page 391 and 392: 11.12: a) b) x = 6 .25 m when FA =
Page 393 and 394: 11.18: a) Denote the length of the
Page 395 and 396: 11.30: a) The volume would increase
Page 397 and 398: 11.43: For the airplane to remain i
Page 399 and 400: 11.49: The horizontal component of
Page 401 and 402: 11.56: (a) and (b) Lower rod: Στ
Page 403 and 404: 11.59: a) ∑τ = 0, axis at lower
Page 405 and 406: −4 −4 2 11.64: a) ∆w = −σ
Page 407 and 408: 11.68: (a) ΣτElbow = 0 F (3.80 cm
Page 409 and 410: 11.71: a) Consider the forces on th
Page 411 and 412: 11.74: a) The center of gravity of
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11.79: a) Take torques about the po
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11.87: Use subscripts 1 to denote t
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11.93: a) Taking torques about the
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11.95: Assume that the center of gr
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