MCQs / Physics / Circular and Rotational Motion

Circular and Rotational Motion MCQs — Physics (page 7)

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.

151. A wheel turns at 120 revolutions per minute. Its angular velocity is:

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

Answer: A) 12.6 rad/s

Explanation: omega = 2pi f = 2pi 12060 = 12.6 rad/s. SLO P-11-B-08 · Understanding

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152. In a rotating space station, the 'floor' is:

  1. the inner hub
  2. the outer rim
  3. the axis of rotation
  4. any wall

Answer: B) the outer rim

Explanation: Astronauts are pressed against the outer wall. SLO P-11-B-17 · Understanding

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153. In a blood centrifuge, the red blood cells collect:

  1. evenly throughout
  2. at the bottom (outer end) of the tube
  3. at the top (near the axis)
  4. outside the tube

Answer: B) at the bottom (outer end) of the tube

Explanation: They are the densest component. SLO P-11-B-21 · Application

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154. A 0.05 kg mass rotates at 30 rad/s on a radius of 0.1 m. The centripetal force is:

  1. 0.15 N
  2. 4.5 N
  3. 9 N
  4. 2.25 N

Answer: B) 4.5 N

Explanation: F = mromega^2 = 0.05 0.1 30^2 = 4.5 N. SLO P-11-B-14 · Application

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155. An ice skater with I = 5 kg m^2 spins at 1 rad/s, then pulls in her arms so I = 2.5 kg m^2. Her new angular velocity is:

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

Answer: B) 2 rad/s

Explanation: I1omega1 = I2omega2 Rightarrow omega2 = 5 12.5 = 2 rad/s. SLO P-11-B-20 · Application

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156. Uranium enrichment uses gas centrifuges to separate:

  1. liquids from gases
  2. electrons from nuclei
  3. isotopes of slightly different mass
  4. colours of light

Answer: C) isotopes of slightly different mass

Explanation: Heavier molecules move slightly further out. SLO P-11-B-21 · Application

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157. A wheel starts at 30 rad/s with angular acceleration -3 rad/s^2. After 8 s its angular velocity is:

  1. 3 rad/s
  2. 6 rad/s
  3. -23 rad/s
  4. 12 rad/s

Answer: B) 6 rad/s

Explanation: omegaf = omegai + alpha t = 30 + (-3)(8) = 6 rad/s. SLO P-11-B-09 · Application

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158. A disc turns through 200 rad in 10 s starting from rest with uniform alpha. Its angular acceleration is:

  1. 2 rad/s^2
  2. 4 rad/s^2
  3. 20 rad/s^2
  4. 40 rad/s^2

Answer: B) 4 rad/s^2

Explanation: 200 = 12alpha(10)^2 Rightarrow alpha = 4 rad/s^2. SLO P-11-B-09 · Application

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159. A wheel turns at 1500 revolutions per minute. Its angular velocity is:

  1. 157 rad/s
  2. 78.5 rad/s
  3. 314 rad/s
  4. 25 rad/s

Answer: A) 157 rad/s

Explanation: omega = 2pi f = 2pi 150060 = 157 rad/s. SLO P-11-B-08 · Understanding

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160. A point is 2 m from the axis of a wheel rotating at 3 rad/s. Its linear speed is:

  1. 3 m/s
  2. 6 m/s
  3. 18 m/s
  4. 1.5 m/s

Answer: B) 6 m/s

Explanation: v = romega = (2)(3) = 6 m/s. SLO P-11-B-08 · Understanding

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161. The work done by a force perpendicular to the velocity is:

  1. negative
  2. zero
  3. equal to Fv
  4. maximum

Answer: B) zero

Explanation: W = Fdcos 90^circ = 0. SLO P-11-B-10 · Understanding

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162. The rotational analogue of Newton's second law is:

  1. L = Iomega
  2. F = ma
  3. tau = Frcostheta
  4. tau = Ialpha

Answer: D) tau = Ialpha

Explanation: Torque = moment of inertia × angular acceleration. SLO P-11-B-19 · Application

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163. The spin-dryer of a washing machine works like a centrifuge because:

  1. water passes through the holes as the drum can't hold it in a circle
  2. air pushes water out
  3. clothes absorb more water
  4. water is evaporated

Answer: A) water passes through the holes as the drum can't hold it in a circle

Explanation: No inward force on the water at the holes. SLO P-11-B-21 · Application

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164. A 0.1 kg ball on a 0.5 m string can withstand a tension of 20 N. The greatest speed at which it can be whirled in a horizontal circle (ignore gravity) is:

  1. 10 m/s
  2. 100 m/s
  3. 5 m/s
  4. 10 m/s

Answer: A) 10 m/s

Explanation: T = mv^2r Rightarrow v = Trm = 20 0.50.1 = 10 m/s. SLO P-11-B-14 · Application

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165. A space station is a ring of radius 250 m. To produce artificial gravity of 9.8 m/s^2 at the rim, it must rotate at:

  1. 0.099 rad/s
  2. 0.0392 rad/s
  3. 0.198 rad/s
  4. 0.396 rad/s

Answer: C) 0.198 rad/s

Explanation: romega^2 = g Rightarrow omega = g/r = 9.8/250 = 0.198 rad/s. SLO P-11-B-17 · Understanding

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166. For an electron orbiting a nucleus (Bohr model), the centripetal force is:

  1. the magnetic force
  2. the electrostatic attraction
  3. friction
  4. gravity

Answer: B) the electrostatic attraction

Explanation: Coulomb attraction pulls it inward. SLO P-11-B-15 · Application

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167. A 1000 kg body moves at 20 m/s in a circle of radius 50 m. The centripetal force is:

  1. 8000 N
  2. 400 N
  3. 16,000 N
  4. 4000 N

Answer: A) 8000 N

Explanation: F = mv^2r = 1000 20^250 = 8000 N. SLO P-11-B-14 · Application

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168. A rotating space station of radius 50 m gives g = 9.8 m/s^2 at the rim. Its period of rotation is:

  1. 14.3 s
  2. 7.1 s
  3. 14.2 s
  4. 28.4 s

Answer: C) 14.2 s

Explanation: omega = 0.443 rad/s, so T = 2pi/omega = 14.2 s. SLO P-11-B-17 · Understanding

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169. To get the same artificial gravity in a station twice as large (radius doubled), the angular speed must be:

  1. halved
  2. unchanged
  3. divided by 2
  4. doubled

Answer: C) divided by 2

Explanation: omega = g/r. SLO P-11-B-17 · Understanding

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170. A rotating space station of radius 100 m gives g = 9.8 m/s^2 at the rim. Its period of rotation is:

  1. 10 s
  2. 20.1 s
  3. 20.2 s
  4. 40.1 s

Answer: B) 20.1 s

Explanation: omega = 0.313 rad/s, so T = 2pi/omega = 20.1 s. SLO P-11-B-17 · Understanding

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171. For a satellite in circular orbit, the centripetal force is provided by:

  1. air resistance
  2. the gravitational force of the Earth
  3. tension
  4. the rocket engine

Answer: B) the gravitational force of the Earth

Explanation: Gravity points to the Earth's centre. SLO P-11-B-15 · Application

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172. At the top of a vertical loop at minimum speed, the normal force from the track is:

  1. twice the weight
  2. infinite
  3. zero
  4. equal to the weight

Answer: C) zero

Explanation: Weight alone provides the centripetal force. SLO P-11-B-15 · Application

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173. A rotating space station of radius 250 m gives g = 9.8 m/s^2 at the rim. Its period of rotation is:

  1. 31.7 s
  2. 63.5 s
  3. 31.8 s
  4. 15.9 s

Answer: A) 31.7 s

Explanation: omega = 0.198 rad/s, so T = 2pi/omega = 31.7 s. SLO P-11-B-17 · Understanding

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174. Water poured from a bottle in an orbiting spacecraft would:

  1. freeze immediately
  2. flow upward
  3. float as blobs instead of falling to the floor
  4. fall normally

Answer: C) float as blobs instead of falling to the floor

Explanation: Everything is in free fall together. SLO P-11-B-16 · Understanding

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175. A space station is a ring of radius 50 m. To produce artificial gravity of 9.8 m/s^2 at the rim, it must rotate at:

  1. 0.885 rad/s
  2. 0.221 rad/s
  3. 0.196 rad/s
  4. 0.443 rad/s

Answer: D) 0.443 rad/s

Explanation: romega^2 = g Rightarrow omega = g/r = 9.8/50 = 0.443 rad/s. SLO P-11-B-17 · Understanding

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