Unit 5 Test Flashcards by dhruv patel

(A) 6 radians
(B) 12 radians
(C) 18 radians
(D) 24 radians
(E) 48 radians

(D) 24 radians

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(A) 1 Nm
(B) 2 Nm
(C) 2.5 Nm
(D) 7 Nm
(E) 7.5 N*m

(A) 1 N*m

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(A) (FRT^2) / 2M
(B) (FRT^2) / 2I
(C) (FT^2) / 2I
(D) FT / M
(E) FRT / I

(B) (FRT^2) / 2I

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(A) 1.1 rad/s
(B) 3.9 rad/s
(C) 5.0 rad/s
(D) 8.9 rad/s
(E) 16 rad/s

(D) 8.9 rad/s

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(A) 2 radians * s–1
(B) 4 radians * s–1
(C) 8 radians * s–1
(D) 16 radians * s–1
(E) 32 radians * s–1

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(A) sqrt(2a)
(B) sqrt(2pia)
(C) sqrt(4pia)
(D) 2a
(E) 4pia

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(A) a = 1/3g
(B) a = 1/2g
(C) a = 2/5g
(D) a = 3/5g
(E) a = g

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(A) w/T
(B) w/T^2
(C) Iw^2 / T
(D) Iw / T^2
(E) Iw / T

(E) Iw / T

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(A) MwR
(B) Mw^2 R
(C) MwR^2
(D) (Mw^2 R^2) / 2
(E) Zero

(A) MwR

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(A) m1 = m2
(B) am1 = bm2
(C) am2 = bm1
(D) a^2 m1 = b^2 m2
(E) b^2 m1 = a^2 m2

(B) am1 = bm2

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(A) T1 = T2 because the wheel has mass
(B) T1 = T2 because both blocks have the same acceleration
(C) T1 > T2 because m1 is farther from the wheel than m2
(D) T1 > T2 because m1 accelerates upwards
(E) T2 > T1 because an unbalanced clockwise torque is needed to accelerate the wheel clockwise

(E) T2 > T1 because an unbalanced clockwise torque is needed to accelerate the wheel clockwise

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(A) m2gR = Ia
(B) (T1 + T2)R = Ia
(C) T2R = Ia
(D) (T2 - T1)R = Ia
(E) (m2 - m1)gR = Ia

(D) (T2 - T1)R = Ia

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(A) From t = 0 s to t = 1 s
(B) From t = 1 s to t = 3 s
(C) From t = 3 s to t = 5 s
(D) From t = 5 s to t = 7 s
(E) From t = 7 s to t = 8 s

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(A) u >= mg
(B) u >= g / (w^2 R)
(C) u >= (w^2 R) / g
(D) u <= g / (w^2 R)
(E) u <= (w^2 R) / g

(B) u >= g / (w^2 R)

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(A) rB
(B) rsqrt(B^2 + C^4)
(C) rsqrt(B^2 + (B + C)^4)
(D) rsqrt((B + C)^2 + C^4)

(B) rsqrt(B^2 + C^4)

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(A) It has magnitude R0w^2 and is directed toward the center of the platform
(B) It has a magnitude R0a0 and is directed tangent to the coin’s circular path
(C) It has a magnitude R0 sqrt(w^4 + a^2). It is initially directed toward the center of the platform, with an increasing component tangent to the coin’s circular path
(D) It has a magnitude R0 sqrt(w^4 + a0^2). It is initially tangent to the coin’s circular path, with an increasing component directed toward the center of the platform.

(D) It has a magnitude R0 sqrt(w^4 + a0^2). It is initially tangent to the coin’s circular path, with an increasing component directed toward the center of the platform.

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(A) 12At^2
(B) 12r0 At^2
(C) 4At^3 + B
(D) r0(4At^3 + B)^2

(B) 12r0 At^2

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(A) vA > vB
(B) vA < vB
(C) vA = vB > 0
(D) vA = vB = 0

(A) vA > vB

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(A) wF = wR, both clockwise
(B) wF = wR, both counterclockwise
(C) wF > wR, both clockwise
(D) wF > wR, both counterclockwise

(D) wF > wR, both counterclockwise

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(A) Multiply the given rotation rate by 2pi and then multiply by the given radius
(B) Multiply the given rotation rate by 2pi and then divide by the given radius
(C) Multiply the given rotation rate by the given radius.
(D) Divide the given rotation rate by the given radius.

(A) Multiply the given rotation rate by 2pi and then multiply by the given radius

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(A) The linear and angular speeds are both the same.
(B) The linear and angular speeds are both different.
(C) The linear speeds are the same, and the angular speeds are different.
(D) The linear speeds are different, and the angular speeds are the same.

(D) The linear speeds are different, and the angular speeds are the same.

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(A) 7.5 kg
(B) 10 kg
(C) 30 kg
(D) 160 kg
(E) 225 kg

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(A) mg
(B) mw^2 r^2 + mg
(C) mw^2 r^2 - mg
(D) mwr^2 - mg
(E) mw^2 r + mg

(E) mw^2 r + mg

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(A) zero
(B) FR
(C) 2FR
(D) 5FR
(E) 14FR

(D) 5FR

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(A) Tension in I Tension in II 75 N 225 N (B) Tension in I Tension in II 100 N 200 N (C) Tension in I Tension in II 112.5 N 112.5 N (D) Tension in I Tension in II 112.5 N 187.5 N (E) Tension in I Tension in II 150 N 300 N

(B) Tension in I Tension in II 100 N 200 N

(A) 2L / 9 (B) 2L / 3 (C) L / 3 (D) L / 2 (E) 3L / 2

(A) 2L / 9

(A) M1 + M2 (B) (M1 + M2) / 2 (C) M1M2 (D) (M1M2) / 2 (E) sqrt(M1M2)

(E) sqrt(M1M2)

(A) 1 / (3A) (B) 1/ (2A) (C) 2A (D) 3A

(D) 3A

(A) 4 radians (B) 16 radians (C) 20 radians (D) 24 radians (E) 36 radians

(E) 36 radians

(A) w0 / (1 + ATw0) (B) w0 e^(-AT) (C) w0( 1 - e^(-AT)) (D) w0 + AT (E) w0 - Aw^2 T

(A) w0 / (1 + ATw0)

(A) 4 rad/s^2 clockwise (B) 4 rad/s^2 counterclockwise (C) 8 rad/s^2 clockwise (D) 8 rad/s^2 counterclockwise

(D) 8 rad/s^2 counterclockwise

(A) w(t) = 12/5 t^(5/2) - t^2 + 2pit (B) w(t) = 9t^(1/2) - 2 (C) w(t) = 6t^(1/2) - 2 (D) w(t) = 9/2t^(-1/2)

(B) w(t) = 9t^(1/2) - 2

(A) because Disk A has a greater angular velocity at the end of the time interval shown. (B) because neither curve has a decreasing slope. (C) because Disk B stops rotating after 1 second. (D) because the area under Disk ’s A curve is greater than the area under Disk ’s B curve.

(D) because the area under Disk ’s A curve is greater than the area under Disk ’s B curve.

(A) Evaluate w(2) (B) Evaluate w(1) and w(3). Take the average of the two values (C) Evaluate the deriviative dw/dt at t = 2. Multiply the result by t = 2 (D) integrate ω(t) from 1 to 3, divide by t = 2 s.

(D) integrate ω(t) from 1 to 3, divide by Δt = 2 s.

(A) A (B) B (C) C (D) D

(A) A

(A) 4θ₁ (B) 2θ₁ (C) 1/2 θ₁ (D) 1/4 θ₁

(B) 2θ₁

(A) αB = ½αA (B) αB = 1/sqrt(2) αA (C) αB = αA (D) αB = 2αA

(A) αB = ½αA

(A) ½βRt₁² (B) ½βt₁² (C) βR (D) β

(A) ½βRt₁²

(A) 1.0 rad/s (B) 1.4 rad/s (C) 2.4 rad/s (D) 3.0 rad/s

(B) 1.4 rad/s

(A) A (B) B (C) C (D) D

(A) A

(A) It is at its maximum magnitude and in the counterclockwise direction. (B) It is at its maximum magnitude and in the clockwise direction. (C) It is increasing in magnitude. (D) It is decreasing in magnitude.

(A) — ω is at its maximum magnitude and in the counterclockwise direction.

(A) θ = A + Bt + 2Ct² (B) θ = A + 2Bt + Ct² (C) θ = A + 2Bt + 2Ct² (D) θ = 2A + 2Bt + 2Ct²

(A) 1 rad/s (B) 2 rad/s (C) 3 rad/s (D) 4 rad/s

(A) 2.5 N (B) 4.3 N (C) 5.0 N (D) 8.7 N

(B) 4.3 N

(A) 0.5 rad/s (B) 1 rad/s (C) 2 rad/s (D) 3 rad/s (E) 4 rad/s

(E) 4 rad/s

(A) W / 2cosθ (B) W / 2sinθ (C) W / cosθ (D) W / sinθ (E) W

(B): W/2sinθ

(A) 48 N (B) 64 N (C) 72 N (D) 120 N (E) 144 N

(A): 48 N

(A) 6 N·m (B) 10 N·m (C) 30 N·m (D) 60 N·m (E) 70 N·m

(A) 6 N·m

(A) 25 N·m (B) 20 N·m (C) 15 N·m (D) 5 N·m

(B) 20 N·m

(A) RF₀/Is (B) 2RF₀/Is (C) 3RF₀/Is (D) 6RF₀/Is

(D) 6RF₀/Is

(A) Point Q. The rope is shorter when attached to Point Q. (B) Point Q. The lever arm is shorter when the rope is attached to Point Q. (C) Point R. The rope is longer when attached to Point R. (D) Point R. The lever arm is longer when the rope is attached to Point R.

(B) — Point Q has greater tension because its shorter lever arm requires more tension to produce the same torque.

(A) 250 g at the center (50 cm) (B) 250 g at ~75 cm (C) 250 g at ~25 cm and 250 g at ~75 cm (D) 300 g at ~50 cm (E) 500 g at ~75 cm

(E) — the largest mass placed at the greatest distance produces the largest torque.

(A) 0 (B) 6 N·m (C) 10 N·m (D) 18 N·m

(B) 6 N·m

(A) F₁, because the lever arm is zero for F₂. (B) F₁, because the product F₁L is greater than F₂(L/3). (C) F₂, because F₂ is greater than F₁. (D) F₂, because F₂ is directed along the length of the rod.

(A) — F₁ produces greater torque because the lever arm for F₂ is zero (it acts along the rod's length).

(A) a (B) b (C) c (D) d (E) The spool will roll on the table for all four directions shown.

(D) d

(A) τ₃ > τ₂ > τ₁ (B) τ₂ > τ₁ = τ₃ (C) τ₁ > τ₂ > τ₃ (D) τ₁ = τ₃ > τ₂

(B) τ₂ > τ₁ = τ₃

(A) Δτ₁₀ = Δτ₂₁ = 0 (B) Δτ₁₀ = Δτ₂₁ > 0 (C) Δτ₁₀ < Δτ₂₁ (D) Δτ₁₀ > Δτ₂₁

(D): Δτ₁₀ > Δτ₂₁

(A) τ₁ = 0 N·m and τ₂ = 10 N·m (B) τ₁ = 0 N·m and τ₂ = 20 N·m (C) τ₁ = 10 N·m and τ₂ = 10 N·m (D) τ₁ = 10 N·m and τ₂ = 20 N·m

(A) τ₁ = 0 N·m and τ₂ = 10 N·m

(A) Only Student 1 is correct. Student 2 does not account for the angle of the force. (B) Only Student 1 is correct. The force is exerted at a distance L₀, not r⊥, from the pivot. (C) Both students are correct. The line of action is parallel to the force. (D) Both students are correct. The perpendicular distance r⊥ is equivalent to L₀sinθ₀.

(D) Both students are correct. The perpendicular distance r⊥ is equivalent to L₀sinθ₀.

(A) 2F₀ > F₀ (B) 2F₀ > F₀ and 2r₀ > r₀ (C) 2F₀ > F₀ and 2L₀ > L₀ (D) 2F₀ > F₀, 2r₀ > r₀, and 2L₀ > L₀

(B) 2F₀ > F₀ and 2r₀ > r₀

(A) 30 N·m (B) 40 N·m (C) 50 N·m (D) 100 N·m

(B) 40 N·m

(A) Both have same angular speed, but B has greater tangential speed. (B) Both have same tangential speed, but B has greater angular speed. (C) Both have the same angular speed and tangential speed. (D) Both have the same angular speed, but A has greater tangential speed. (E) A has both greater angular speed and greater tangential speed.

(D) Both have the same angular speed, but A has greater tangential speed.

(A) 4 (B) 2 (C) 1 (D) 1/2 (E) 1/4

(A) ¼ · kT²/π² (B) ¾ · kT²/π² (C) 3/2 · kT²/π² (D) 3 · kT²/π²

(B) ¾ · kT²/π²

(A) Net torque is required to turn the pulley with mass, so F₁ and F₂ both decrease. (B) Net torque is required to turn the pulley with mass, so F₁ and F₂ both increase. (C) Net torque is required to turn the pulley with mass, so F₁ increases and F₂ decreases. (D) Net torque is required to turn the pulley with mass, so F₁ decreases and F₂ increases. (E) The angular acceleration α was incorrectly determined, because increasing the mass of the pulley should not affect α.

(D) Net torque is required to turn the pulley with mass, so F₁ decreases and F₂ increases.

(A) Segment A (B) Segment B (C) Segment C (D) Segment D (E) Segment E

(A) Segment A

(A) ω/4 (B) ω/2 (C) ω (D) 2ω (E) 4ω

(E) 4ω

The sphere-rod combination can be pivoted about an axis that is perpendicular to the plane of the page and that passes through one of the five lettered points. Through which point should the axis pass for the moment of inertia of the sphere-rod combination about this axis to be greatest? (A) A (B) B (C) C (D) D (E) E

(E): E

(A) ⅓ω₀ (B) ½ω₀ (C) ⅔ω₀ (D) 4/5 ω₀ (E) 2/√5 ω₀

(D) 4/5 ω₀

(A) The ring will move farther than will the disk. (B) The disk will move farther than will the ring. (C) The ring and the disk will move equal distances. (D) The relative distances depend on the angle of elevation of the plane. (E) The relative distances depend on the length of the plane.

(A) The ring will move farther than will the disk.

(A) (T − mg)/m (B) 2g (C) g/2 (D) T/m (E) zero

(B) 2g

(A) mbt/r (B) mbrt (C) mbr²t (D) mbr²rt² (E) mb²r²t²

(A) ⅓ mℓ² (B) ⅙ mℓ² (C) 1/12 mℓ² (D) 1/48 mℓ² (E) 7/48 mℓ²

(E) 7/48 mℓ²

(A) Hollow cylinder, hollow sphere, solid cylinder, solid sphere (B) Hollow cylinder, solid sphere, solid cylinder, hollow sphere (C) Solid sphere, solid cylinder, hollow sphere, hollow cylinder (D) Solid sphere, hollow sphere, solid cylinder, hollow cylinder (E) Hollow sphere, solid cylinder, hollow cylinder, solid sphere

(A) 1/2 (B) 1/√2 (C) √2 (D) 2

(A) 2/1 (B) 1/1 (C) 1/2 (D) 1/4 (E) 1/8

(E) 1/8

(A) Replace with a ring of same mass and radius. (B) Replace with a disk of same mass and radius but density decreasing radially. (C) Replace with a disk of twice the radius but same mass. (D) Exert the force perpendicular to the edge rather than tangent.

(B) Replace with a disk of same mass and radius but density decreasing radially.

(A) 3g/2L (B) g(L + 2x) / (⅓L² + 2x²) (C) g(½L + x) / (⅓L² + x²) (D) g(½L + 2x) / (⅓L² + 2x²)

(D) g(½L + 2x) / (⅓L² + 2x²)

(A) M₁g(⅓L_B) − M₂g(⅔L_B) = 0 (B) M₁g(⅓L_B) − M_Bg(½L_B) − M₂g(⅔L_B) = 0 (C) M₁g(⅓L_B) + M_Bg(⅙L_B) − M₂g(⅔L_B) = 0 (D) M₁g(⅓L_B) − M_Bg(⅙L_B) − M₂g(⅔L_B) = 0

(D) M₁g(⅓L_B) − M_Bg(⅙L_B) − M₂g(⅔L_B) = 0

(A) 3/20 M₀ (B) 3/10 M₀ (C) 3/8 M₀ (D) 1/2 M₀

(A) α₁ > α₂ = 0 (B) α₁ > α₂ ≠ 0 (C) α₂ > α₁ = 0 (D) α₂ > α₁ ≠ 0

(A) α₁ > α₂ = 0

(A) ω increases (concave up) then becomes constant after t₁ (B) ω increases linearly then becomes constant after t₁ (C) ω increases concave up continuously past t₁ (D) ω increases concave up with no flattening

(B) ω increases linearly then becomes constant after t₁

(A) m₀g (B) √2 m₀g (C) √17 m₀g (D) √32 m₀g

(A) F₁ = F₂ = F₃ (B) F₁ > F₂ > F₃ (C) F₂ = F₃ > F₁ (D) F₂ > F₃ > F₁

(D): F₂ > F₃ > F₁

(A) Between t = 2s and t = 3s only (B) Between t = 2s and t = 3s, and between t = 4s and t = 5s (C) Between t = 4s and t = 5s only (D) At no time between 0 and 5 seconds is the angular velocity constant.

(A) Zero (B) m₀s₀² (C) 2m₀s₀² (D) 4m₀s₀²

(B) m₀s₀²

(A) ¼ I₁ (B) ½ I₁ (C) ¾ I₁ (D) 3/2 I₁

(D) 3/2 I₁

(A) Rod A (B) Rod B (C) Rod C (D) Rods A, B, and C have the same rotational inertia.

(B) Rod B

(A) 1/2 (B) 1/√2 (C) √2 (D) 2

(A) 1/4 (B) 1/3 (C) 1/2 (D) 7/12

(A) 1/4

(A) (1/3)βh₀³ (B) (4/3)βh₀³ (C) (1/4)βh₀⁴ (D) (1/5)βh₀⁵

(D) (1/5)βh₀⁵

(A) 8 kg·m² (B) 13 kg·m² (C) 32 kg·m² (D) 53 kg·m² (E) 80 kg·m²

(A) (1/36)mL² (B) (7/36)mL² (C) (1/4)mL² (D) (5/12)mL² (E) (3/4)mL²

(B) (7/36)mL²

(A) (3/2)MR² (B) (7/5)MR² (C) MR² (D) (1/2)MR² (E) (2/3)MR²

(B) (7/5)MR²

(A) 2MR² (B) 3MR² (C) 4MR² (D) 5MR² (E) 10MR²

(B) 3MR²

(A) I_X > I_Y > I_Z (B) I_X > I_Z > I_Y (C) I_Z > I_X > I_Y (D) I_Z > I_Y > I_X

(B) I_X > I_Z > I_Y

(A) System A (B) System B (C) System C (D) All three systems have the same rotational inertia

(D) All three systems have the same rotational inertia

(A) 1/3 (B) 1/2 (C) 2/3 (D) 3/2

(A) Disk A will hit the ground first. Disk A has a greater rotational inertia; therefore, the tension in the string for Disk A will be greater than the tension in the string for Disk B. (B) Disk B will hit the ground first. Disk B has a smaller rotational inertia; therefore, it will have a greater angular acceleration. (C) Disk A will hit the ground first. Disk A has a smaller rotational inertia; therefore, it will have a greater linear acceleration. (D) Disk B will hit the ground first. Disk B has a greater rotational inertia; therefore, it will have a greater angular acceleration. (E) Both disks will hit the ground at the same time. Since both disks have the same mass, they will have the same linear acceleration.

(C) Disk A will hit the ground first. Disk A has a smaller rotational inertia; therefore, it will have a greater linear acceleration.

(A) The radius of the hinge is not negligible. (B) The linear density of the short cable increases vertically. (C) The linear density of the bar decreases with distance from the wall. (D) The rotational inertia of the bar about the hinge is less than (1/3)ML². (E) The rotational inertia of the bar about the hinge is greater than (1/3)ML².

(E) The rotational inertia of the bar about the hinge is greater than (1/3)ML².

(A) wf = ( I / ( I + StR)) w0 (B) wf = ( I / ( I + SR)) w0 (C) wf = ( I / ( I + StR^2)) w0 (D) wf = (2I) / ((I + StR^2)t) w0 (E) wf = (2I) / ((I + StR^2)t) w0

(A) 2 : 1 (B) 1 : 2 (C) 1 : 4 (D) 1 : 8

(A) 5/4 g/L0 (B) 3/2 g/L0 (C) 2 g/L0 (D) 5/2 g/L0

(A) 5/4 g/L0

(A) 1 (B) 4/5 (C) 3/2 (D) 2/5

(B) 4/5

(A) 5a0 (B) 3a0 (C) 5/3a0 (D) a0

(A) 5a0

(A) a = 0. The angular acceleration cannot be zero because the two torques about the rod's center of mass are in the same direction (B) a = 0. The angular acceleration cannot be zero because the distances from the center of mass to the point at which each force is exerted is different (C) a not equal to 0. The angular acceleration cannot be zero because the two torques about the rod's center of mass are in the same direction (D) a not equal to 0. The angular acceleration cannot be zero because the distances from the center of mass to the point at which each force is exerted are different.

(A) a = 0. The angular acceleration cannot be zero because the two torques about the rod's center of mass are in the same direction

(A) It increases. The area under the curve increases with time. (B) It increases. The second derivative of the curve is positive. (C) It decreases. The slope of the curve decreases as time increases. (D) It decreases. The angular speed value decreases as time increases.

(A) 1/11 g/D0 (B) 1/7 g/D0 (C) 1/3 g/D0 (D) 5/11 g/D0

(A) 1/11 g/D0

Unit 5 Test Flashcards

(108 cards)