MCQs / Physics / Circular and Rotational Motion

Circular and Rotational Motion MCQs — Physics (page 5)

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.

101. The angular equations of motion are valid only for:

  1. zero angular velocity
  2. constant angular acceleration
  3. variable angular acceleration
  4. constant linear speed

Answer: B) constant angular acceleration

Explanation: Like the linear equations. SLO P-11-B-09 · Application

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102. A bucket of water is swung in a vertical circle of radius 2.5 m. The minimum speed at the top so the water stays in is (g = 9.8 m/s^2):

  1. 9.9 m/s
  2. 2.47 m/s
  3. 7 m/s
  4. 4.95 m/s

Answer: D) 4.95 m/s

Explanation: At the minimum speed, weight alone provides the centripetal force: mg = mv^2r Rightarrow v = gr = 4.95 m/s. SLO P-11-B-15 · Application

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103. Which balance must be kept in a laboratory centrifuge?

  1. tubes of different masses
  2. all tubes on one side
  3. one tube only
  4. tubes of equal mass placed opposite each other

Answer: D) tubes of equal mass placed opposite each other

Explanation: Unbalanced rotation causes vibration and damage. SLO P-11-B-21 · Application

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104. During the free-fall part of a 'zero-g' aircraft flight, passengers feel weightless because:

  1. gravity is switched off
  2. the aircraft follows a free-fall (parabolic) path
  3. the engines push them up
  4. the aircraft leaves the atmosphere

Answer: B) the aircraft follows a free-fall (parabolic) path

Explanation: Same free-fall idea as in orbit. SLO P-11-B-16 · Understanding

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105. A centrifuge tube is 0.1 m from the axis and spins at 3000 rpm. The centripetal acceleration is about how many times g (g = 9.8)?

  1. 2010 g
  2. 504 g
  3. 3.21 g
  4. 1010 g

Answer: D) 1010 g

Explanation: omega = 314 rad/s; a = romega^2 = 9870 m/s^2 approx 1010g. SLO P-11-B-21 · Application

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106. A 0.2 kg mass rotates at 10 rad/s on a radius of 0.5 m. The centripetal force is:

  1. 5 N
  2. 1 N
  3. 10 N
  4. 20 N

Answer: C) 10 N

Explanation: F = mromega^2 = 0.2 0.5 10^2 = 10 N. SLO P-11-B-14 · Application

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107. 90^circ expressed in radians is:

  1. pi4
  2. pi
  3. pi2
  4. 2pi

Answer: C) pi2

Explanation: Multiply by pi180^circ: 90 pi180 = pi2 rad. SLO P-11-B-07 · Understanding

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108. An arc of length 6 m on a circle of radius 2 m subtends an angle of:

  1. 1.5 rad
  2. 172 rad
  3. 3 rad
  4. 6 rad

Answer: C) 3 rad

Explanation: theta = sr = 62 = 3 rad. SLO P-11-B-07 · Understanding

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109. A bathroom scale in an orbiting spacecraft would read:

  1. a negative value
  2. twice the weight
  3. zero
  4. the astronaut's normal weight

Answer: C) zero

Explanation: The scale and astronaut fall together, so there is no contact force. SLO P-11-B-16 · Understanding

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110. Moment of inertia plays the same role in rotation as:

  1. mass plays in linear motion
  2. energy
  3. force
  4. velocity

Answer: A) mass plays in linear motion

Explanation: It measures resistance to angular acceleration. SLO P-11-B-18 · Application

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111. Centripetal force is:

  1. not a new kind of force — it is the name for the net inward force
  2. a force that pushes outward
  3. always gravitational
  4. a fictitious force

Answer: A) not a new kind of force — it is the name for the net inward force

Explanation: Tension, friction, gravity etc. can provide it. SLO P-11-B-14 · Application

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112. The quantity 'relative centrifugal force (RCF)' compares:

  1. speed with time
  2. romega^2 with g
  3. density with volume
  4. mass with weight

Answer: B) romega^2 with g

Explanation: RCF = romega^2/g. SLO P-11-B-21 · Application

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113. A wheel starts at 30 rad/s with angular acceleration -3 rad/s^2. In 8 s it turns through:

  1. 72 rad
  2. 288 rad
  3. 48 rad
  4. 144 rad

Answer: D) 144 rad

Explanation: theta = omegai t + 12alpha t^2 = 144 rad (about 22.9 revolutions). SLO P-11-B-09 · Application

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114. Torque is maximum when a force is applied:

  1. perpendicular to the lever arm, far from the axis
  2. parallel to the axis
  3. at the axis
  4. along the lever arm

Answer: A) perpendicular to the lever arm, far from the axis

Explanation: tau = rFsintheta. SLO P-11-B-19 · Application

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115. In orbit, the force of gravity on an astronaut:

  1. is balanced by a centrifugal force
  2. pushes the astronaut outward
  3. is zero
  4. provides the centripetal force for circular motion

Answer: D) provides the centripetal force for circular motion

Explanation: Gravity is the only force, and it curves the path. SLO P-11-B-16 · Understanding

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116. A force of 20 N is applied tangentially at the rim (radius 0.5 m) of a wheel with I = 2 kg m^2. Its angular acceleration is:

  1. 2.5 rad/s^2
  2. 10 rad/s^2
  3. 5 rad/s^2
  4. 20 rad/s^2

Answer: C) 5 rad/s^2

Explanation: tau = Fr = 10 N m; alpha = tau/I = 5 rad/s^2. SLO P-11-B-19 · Application

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117. A 0.5 kg body moves at 4 m/s in a circle of radius 0.8 m. The centripetal force is:

  1. 10 N
  2. 2.5 N
  3. 5 N
  4. 20 N

Answer: A) 10 N

Explanation: F = mv^2r = 0.5 4^20.8 = 10 N. SLO P-11-B-14 · Application

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118. A car on a banked road can turn without friction because:

  1. its speed is zero
  2. the horizontal component of the normal force provides the centripetal force
  3. the road is flat
  4. gravity pushes it outward

Answer: B) the horizontal component of the normal force provides the centripetal force

Explanation: tantheta = v^2rg for the ideal bank. SLO P-11-B-15 · Application

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119. 270^circ expressed in radians is:

  1. 2pi3
  2. 3pi4
  3. 3pi
  4. 3pi2

Answer: D) 3pi2

Explanation: Multiply by pi180^circ: 270 pi180 = 3pi2 rad. SLO P-11-B-07 · Understanding

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120. Apparent weight is:

  1. zero on Earth
  2. the mass of a body
  3. the contact (normal) force from the floor or scale
  4. the gravitational force always

Answer: C) the contact (normal) force from the floor or scale

Explanation: We 'feel' weight through support forces. SLO P-11-B-16 · Understanding

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121. A solid disc of mass 10 kg and radius 1 m rotates about its central axis. Its moment of inertia is:

  1. 4 kg m^2
  2. 2.5 kg m^2
  3. 10 kg m^2
  4. 5 kg m^2

Answer: D) 5 kg m^2

Explanation: I = 12MR^2 = 5 kg m^2. SLO P-11-B-18 · Application

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122. For the same mass and radius, which has the larger moment of inertia about its axis?

  1. a solid disc
  2. a ring
  3. they are equal
  4. a solid sphere

Answer: B) a ring

Explanation: All the ring's mass is at the full radius. SLO P-11-B-18 · Application

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123. A flywheel (I = 0.8 kg m^2) is brought from rest to 50 rad/s in 10 s. The torque needed is:

  1. 8 N m
  2. 5 N m
  3. 2 N m
  4. 4 N m

Answer: D) 4 N m

Explanation: alpha = 5 rad/s^2; tau = Ialpha = 4 N m. SLO P-11-B-19 · Application

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124. If the perpendicular force on a body in circular motion suddenly disappears, the body moves:

  1. in a straight line along the tangent
  2. inward to the centre
  3. in a smaller circle
  4. outward along the radius

Answer: A) in a straight line along the tangent

Explanation: With no net force it keeps its velocity (Newton's first law). SLO P-11-B-10 · Understanding

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125. A wheel with moment of inertia 0.5 kg m^2 spins at 20 rad/s. Its angular momentum is:

  1. 10 kg m^2/s
  2. 200 kg m^2/s
  3. 20 kg m^2/s
  4. 5 kg m^2/s

Answer: A) 10 kg m^2/s

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

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