MCQs / Physics / Circular and Rotational Motion

Circular and Rotational Motion MCQs — Physics (page 3)

186 multiple-choice questions on Circular and Rotational Motion from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.

51. A car moves at 20 m/s round a bend of radius 50 m. Its centripetal acceleration is:

  1. 1000 m/s^2
  2. 8 m/s^2
  3. 2.5 m/s^2
  4. 0.4 m/s^2

Answer: B) 8 m/s^2

Explanation: a = v^2/r = 400/50 = 8 m/s^2. SLO P-11-B-08 · Understanding

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52. A point is 0.3 m from the axis of a wheel rotating at 20 rad/s. Its linear speed is:

  1. 6 m/s
  2. 120 m/s
  3. 66.7 m/s
  4. 3 m/s

Answer: A) 6 m/s

Explanation: v = romega = (0.3)(20) = 6 m/s. SLO P-11-B-08 · Understanding

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53. In a washing machine spin cycle, water leaves the clothes because:

  1. water becomes lighter
  2. water is pushed outward by a real force
  3. the drum holes cannot provide centripetal force to the water
  4. gravity increases

Answer: C) the drum holes cannot provide centripetal force to the water

Explanation: Without an inward force the water moves tangentially. SLO P-11-B-15 · Application

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54. A wheel with moment of inertia 4 kg m^2 spins at 1.5 rad/s. Its angular momentum is:

  1. 3 kg m^2/s
  2. 6 kg m^2/s
  3. 12 kg m^2/s
  4. 9 kg m^2/s

Answer: B) 6 kg m^2/s

Explanation: L = Iomega = 6 kg m^2/s. SLO P-11-B-18 · Application

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55. All points on a rotating rigid body have the same:

  1. centripetal acceleration
  2. linear displacement
  3. linear speed
  4. angular velocity

Answer: D) angular velocity

Explanation: v = romega depends on r, but omega is shared. SLO P-11-B-08 · Understanding

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56. A force of 50 N is applied tangentially at the rim (radius 0.2 m) of a wheel with I = 1 kg m^2. Its angular acceleration is:

  1. 5 rad/s^2
  2. 250 rad/s^2
  3. 10 rad/s^2
  4. 20 rad/s^2

Answer: C) 10 rad/s^2

Explanation: tau = Fr = 10 N m; alpha = tau/I = 10 rad/s^2. SLO P-11-B-19 · Application

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57. A centrifuge tube is 0.08 m from the axis and spins at 12000 rpm. The centripetal acceleration is about how many times g (g = 9.8)?

  1. 6450 g
  2. 10.3 g
  3. 12,900 g
  4. 25,800 g

Answer: C) 12,900 g

Explanation: omega = 1260 rad/s; a = romega^2 = 1.26 10^5 m/s^2 approx 12,900g. SLO P-11-B-21 · Application

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58. A diver jumping from a high board feels weightless:

  1. never
  2. while falling freely before entering the water
  3. after entering the water
  4. when standing on the board

Answer: B) while falling freely before entering the water

Explanation: Free fall removes support forces. SLO P-11-B-16 · Understanding

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59. Artificial gravity is strongest:

  1. on the axis
  2. at the centre
  3. everywhere the same
  4. at the outer rim

Answer: D) at the outer rim

Explanation: a = romega^2 grows with r. SLO P-11-B-17 · Understanding

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60. A wheel starts at 0 rad/s with angular acceleration 4 rad/s^2. After 5 s its angular velocity is:

  1. 20 rad/s
  2. 10 rad/s
  3. 40 rad/s
  4. 21 rad/s

Answer: A) 20 rad/s

Explanation: omegaf = omegai + alpha t = 0 + (4)(5) = 20 rad/s. SLO P-11-B-09 · Application

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61. A wheel with moment of inertia 2 kg m^2 spins at 5 rad/s. Its angular momentum is:

  1. 20 kg m^2/s
  2. 50 kg m^2/s
  3. 5 kg m^2/s
  4. 10 kg m^2/s

Answer: D) 10 kg m^2/s

Explanation: L = Iomega = 10 kg m^2/s. SLO P-11-B-18 · Application

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62. 120^circ expressed in radians is:

  1. pi3
  2. 4pi3
  3. 3pi2
  4. 2pi3

Answer: D) 2pi3

Explanation: Multiply by pi180^circ: 120 pi180 = 2pi3 rad. SLO P-11-B-07 · Understanding

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63. Doubling the radius of a centrifuge at the same rpm makes the acceleration:

  1. four times
  2. half
  3. unchanged
  4. double

Answer: D) double

Explanation: a = romega^2 propto r. SLO P-11-B-21 · Application

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64. A disc's angular velocity changes from 0 rad/s to 20 rad/s in 4 s. Its angular acceleration is:

  1. 5 rad/s^2
  2. 2.5 rad/s^2
  3. 10 rad/s^2
  4. -5 rad/s^2

Answer: A) 5 rad/s^2

Explanation: alpha = omegaf - omegait = 5 rad/s^2. SLO P-11-B-08 · Understanding

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65. The angular equations of motion correspond to the linear ones with:

  1. s to theta, v to omega, a to alpha
  2. no correspondence
  3. s to omega, v to theta, a to alpha
  4. s to alpha, v to omega, a to theta

Answer: A) s to theta, v to omega, a to alpha

Explanation: Same form, angular quantities replace linear ones. SLO P-11-B-09 · Application

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66. A wheel starts at 0 rad/s with angular acceleration 4 rad/s^2. In 5 s it turns through:

  1. 25 rad
  2. 100 rad
  3. 52 rad
  4. 50 rad

Answer: D) 50 rad

Explanation: theta = omegai t + 12alpha t^2 = 50 rad (about 7.96 revolutions). SLO P-11-B-09 · Application

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67. A point is 0.3 m from the axis of a wheel rotating at 20 rad/s. Its centripetal acceleration is:

  1. 120 m/s^2
  2. 240 m/s^2
  3. 6 m/s^2
  4. 60 m/s^2

Answer: A) 120 m/s^2

Explanation: a = romega^2 = (0.3)(20)^2 = 120 m/s^2. SLO P-11-B-08 · Understanding

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68. The SI unit of moment of inertia is:

  1. kg m^2
  2. kg m/s
  3. N m
  4. kg/m^2

Answer: A) kg m^2

Explanation: I = mr^2. SLO P-11-B-18 · Application

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69. A solid disc of mass 2 kg and radius 0.5 m rotates about its central axis. Its moment of inertia is:

  1. 0.25 kg m^2
  2. 0.125 kg m^2
  3. 0.2 kg m^2
  4. 0.5 kg m^2

Answer: A) 0.25 kg m^2

Explanation: I = 12MR^2 = 0.25 kg m^2. SLO P-11-B-18 · Application

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70. Deriving tau = Ialpha: for a particle, F = ma = mralpha, so tau = rF =

  1. mr^2alpha
  2. mr^2omega
  3. mralpha
  4. malpha/r

Answer: A) mr^2alpha

Explanation: Summing over all particles gives tau = (sum mr^2)alpha = Ialpha. SLO P-11-B-19 · Application

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71. A wheel turns at 300 revolutions per minute. Its angular velocity is:

  1. 62.8 rad/s
  2. 15.7 rad/s
  3. 31.4 rad/s
  4. 5 rad/s

Answer: C) 31.4 rad/s

Explanation: omega = 2pi f = 2pi 30060 = 31.4 rad/s. SLO P-11-B-08 · Understanding

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72. A body completes one revolution every 0.5 s. Its angular velocity is:

  1. 12.6 rad/s
  2. 25.1 rad/s
  3. 2 rad/s
  4. 6.28 rad/s

Answer: A) 12.6 rad/s

Explanation: omega = 2piT = 2pi0.5 = 12.6 rad/s. SLO P-11-B-08 · Understanding

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73. The formula F = mromega^2 is equivalent to F = mv^2r because:

  1. v = omega
  2. v = omega/r
  3. v = romega
  4. v = r^2omega

Answer: C) v = romega

Explanation: Substitute v = romega: m(romega)^2r = mromega^2. SLO P-11-B-14 · Application

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74. An ice skater with I = 4 kg m^2 spins at 2 rad/s, then pulls in her arms so I = 1 kg m^2. Her new angular velocity is:

  1. 4 rad/s
  2. 8 rad/s
  3. 16 rad/s
  4. 0.5 rad/s

Answer: B) 8 rad/s

Explanation: I1omega1 = I2omega2 Rightarrow omega2 = 4 21 = 8 rad/s. SLO P-11-B-20 · Application

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75. A car rounds a flat bend of radius 50 m. The coefficient of friction is 0.8 (g = 9.8 m/s^2). The maximum safe speed is:

  1. 28 m/s
  2. 9.9 m/s
  3. 39.6 m/s
  4. 19.8 m/s

Answer: D) 19.8 m/s

Explanation: Friction provides the centripetal force: mu mg = mv^2r Rightarrow v = mu g r = 19.8 m/s. SLO P-11-B-15 · Application

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