fisica1-youn-e-freedman-exercicios-resolvidos

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10.101: a) The friction force is f = µ n = µ Mg so a = µ g. The magnitude of the angular acceleration is gives fR I ( 2) k k , k µ k MgR 2 µ k g = = . b) Setting v = at = ωR = ( ω 0 − ωt)R and solving for t 2 1 MR R and Rω0 Rω0 Rω0 t = = = , a + Rα µ g + µ g 3µ g k 2 k k 2 2 1 2 1 ⎛ Rω ⎞ R ω d = at = k 2 2 ⎜ = 3µ kg ⎟ ⎝ ⎠ 18µ kg 0 0 ( µ g) ⎜ ⎟ . 2 is 2 c) The final kinetic energy is ( 3 4) ( 3 4) ( at) , 2 3 ⎛ Rω ⎞ 0 1 2 2 1 2 2 M k MR ω0 MR ω0 . 4 ⎜ µ g − = 3µ kg ⎟ ⎝ ⎠ 4 6 2 Mv = M so the change in kinetic energy
10.102: Denoting the upward forces that the hands exert as F L and F R , the conditions that F L and F R must satisfy are F + F = w L R Iω FL − FR = Ω , r where the second equation is τ = ΩL, divided by r. These two equations can be solved for the forces by first adding and then subtracting, yielding 1 ⎛ Iω ⎞ FL = ⎜ω + Ω ⎟ 2 ⎝ r ⎠ 1 ⎛ Iω ⎞ FR = ⎜ω − Ω ⎟. 2 ⎝ r ⎠ 2 Using the values ω = mg = (8.00 kg)(9.80 m s ) = 78.4 N and gives a) Ω = 0, F 2 Iω (8.00 kg)(0.325 m) (5.00 rev s × 2π rad rev) = = 132.7 kg ⋅ m s r (0.200 m) L F = F 39.2 N + Ω(66.4 N ⋅s), F = 39.2 N − Ω(66.4 N ⋅s). = 39.2 N. L = R R b) Ω = 0.05 rev s = 0.314 rad s, F c) Ω = 0.3 rev s = 1.89 rad s, F d) = 0 gives L = 60.0 N, F = 165 N, F with the minus sign indicating a downward force. F R Ω = 39.2 N = 66.4 N⋅s L R R = 18.4 N. = −86.2 N, 0.575 rad s, which is 0.0916 rev s. 2 2 3 10.103: a) See Problem 10.92; T = mv r . W = − c) v = v ( r ), 2 so 2 1 1 r 1 1 r r r b) T and d are always antiparallel, so 2 r2 r1 2 2 dr mv 2 ⎡ 1 1 ⎤ ∫ T dr = mv r ∫ = 1 r1 ⎢ − 1 3 2 ⎥ ⎦ r1 1 ∆K = m( v 2 2 2 1 . r 2 2 r 2 2 r1 − v 2 1 mv ) = 2 which is the same as the work found in part (b). 2 1 ⎡⎛ r ⎞ 1 ⎢ ⎜ ⎟ ⎢⎣ ⎝ r2 ⎠ 2 ⎣r ⎤ −1⎥ ' ⎥⎦
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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
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8.101: a) If the objects stick toge
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8.107: a) For t < 0 the rocket is a
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8.111: a) The tension in the rope a
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9.1: a) 1.50 m = 0.60 rad = 34.4°
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2 9.10: a) ω = ω + α t = 1.50 ra
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9.18: The following table gives the
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9.24: From a rad = ω 2 r, 4 which
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vr 5.00 m s 9.33: The angular veloc
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9.40: a) In the expression of Eq. (
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2π 2 2 9.49: a) ω = T , so Eq. (9
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9.60: For this case, dm = γ dx. a)
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dθ 2 2 3 2 9.65: a) ω z = = 2γt
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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
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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 and 362: 10.40: The skater’s initial momen
Page 363 and 364: 10.46: (a) Conservation of angular
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: 10.100: a) The distance from the ce
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
Page 413 and 414: 11.79: a) Take torques about the po
Page 415 and 416: 11.87: Use subscripts 1 to denote t
Page 417 and 418: 11.93: a) Taking torques about the
Page 419 and 420: 11.95: Assume that the center of gr
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