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: zero angular velocity constant angular acceleration variable angular acceleration 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): 9.9 m/s 2.47 m/s 7 m/s 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? tubes of different masses all tubes on one side one tube only 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: gravity is switched off the aircraft follows a free-fall (parabolic) path the engines push them up 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)? 2010 g 504 g 3.21 g 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: 5 N 1 N 10 N 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: pi4 pi pi2 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: 1.5 rad 172 rad 3 rad 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: a negative value twice the weight zero 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: mass plays in linear motion energy force 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: not a new kind of force — it is the name for the net inward force a force that pushes outward always gravitational 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: speed with time romega^2 with g density with volume 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: 72 rad 288 rad 48 rad 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: perpendicular to the lever arm, far from the axis parallel to the axis at the axis 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: is balanced by a centrifugal force pushes the astronaut outward is zero 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: 2.5 rad/s^2 10 rad/s^2 5 rad/s^2 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: 10 N 2.5 N 5 N 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: its speed is zero the horizontal component of the normal force provides the centripetal force the road is flat 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: 2pi3 3pi4 3pi 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: zero on Earth the mass of a body the contact (normal) force from the floor or scale 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: 4 kg m^2 2.5 kg m^2 10 kg m^2 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? a solid disc a ring they are equal 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: 8 N m 5 N m 2 N m 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: in a straight line along the tangent inward to the centre in a smaller circle 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: 10 kg m^2/s 200 kg m^2/s 20 kg m^2/s 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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