NJC 2022 Motion in a Circle Problems Set

National Junior College 2022 | SH1 | H2 Physics Motion In a Circle - Problem Set Exercises Kinematics of Uniform Circular Motion E1. What is 𝜋 4 𝑟𝑎𝑑 in degrees? [45°] 𝜋/4 2𝜋 × 360° E2. What is 300° in radians? [5.23 rad] 5 3 𝜋 = 5.23 radians E3. (2020 P1 Q11) [B] 𝜔 = 2𝜋 𝑇 = 2𝜋 60×60 = 1.75 × 10−3 𝑟𝑎𝑑 𝑠−1 E4. (2016 P1 Q12) [D] 𝜔 = 2𝜋 𝑇 = 2𝜋 24×60×60 = 7.27 × 10−5 𝑟𝑎𝑑 𝑠−1

National Junior College 2022 | SH1 | H2 Physics E5. Assuming that the earth moves in a circle at a constant rate around the sun, calculate the angular velocity of the earth around the sun. (Hint: how long does it take for the earth to go round the sun once?) [1.99 × 10−7𝑟𝑎𝑑 𝑠−1] 2𝜋 365×24×60×60 = 1.99 × 10−7𝑟𝑎𝑑 𝑠−1 (The earth takes 365 days to go round the sun. The earth takes 24 hours to rotate about its own axis which results in the day and night cycle.) E6. A car is moving around a circular track of radius 400 𝑚 at constant angular velocity 0.050 𝑟𝑎𝑑 𝑠−1. Calculate the total distance the car moves in 5 minutes. [6000 m] Angular displacement in 5 min = 𝜔𝑡 = 0.05 × 5 × 60 = 15 𝑟𝑎𝑑 Distance moved in 5 min = 𝑟𝜃 = 15 × 400 = 6000 𝑚 E7. A rod is made to spin at a constant rate of 3 complete revolutions per second as shown below. What is the difference in angle between the rod’s current position and its position 0.50 seconds later? [𝜋 𝑟𝑎𝑑] Angular displacement in 0.50 s = 𝜔𝑡 = 3×2𝜋 1 × 0.50 = 3𝜋 The rod would have moved 1.5 complete circles. Hence the difference in angle = 3𝜋 − 2𝜋 = 𝜋 𝑟𝑎𝑑 = 3.14 𝑟𝑎𝑑 E8. (2017 P1 Q9) D (both have the same angular velocity, hence angular displacement will be the same, although actual distance travelled will be larger for P)

National Junior College 2022 | SH1 | H2 Physics Centripetal Force E9. The diagram shows a child sitting on a playground turntable, which is turning with constant angular velocity. Which diagram shows the forces acting on the child when in the position shown? A (centripetal force is a resultant force) E10. (2013 P1 Q11) B (The tension and weight result in a centripetal force to the left pointing to the center of the horizontal circle)

National Junior College 2022 | SH1 | H2 Physics Problems Kinematics of Uniform Circular Motion P1. A body rotates with uniform speed in a circle of radius 𝑟 and takes time 𝑇 to complete one revolution. What are the magnitudes of the angular velocity 𝜔, the linear velocity 𝑣, and the acceleration 𝑎? angular velocity 𝜔 linear velocity 𝑣 acceleration 𝑎 A 1 𝑇 4𝜋𝑟 𝑇 2𝜋𝑟 𝑇2 B 2𝜋 𝑇 2𝜋𝑟 𝑇 2𝜋𝑟 𝑇2 C 2𝜋 𝑇 2𝜋𝑟 𝑇 4𝜋2𝑟 𝑇2 D 2𝜋 𝑇 4𝜋𝑟 𝑇 4𝜋2𝑟 𝑇2 C (𝑣 = 𝑟𝜔, 𝑎 = 𝜔2𝑟) P2. (2017 P1 Q9) D (𝜔 is the same for both points 𝑎 = 𝜔2𝑟 Since radius is halved, the centripetal acceleration is also halved to 8 cm s-2)

National Junior College 2022 | SH1 | H2 Physics P3. (2020 P1 Q10) B (𝑎 = 𝜔2𝑟 = 4𝜋2 (27.3×24×60×60)2 × 3.85 × 108 = 2.73 × 10−3 𝑚𝑠−2) P4. Singapore is on the equator. Cambridge is at a latitude of 52° N, as shown in the figure. A student in Singapore has a centripetal acceleration 𝑎𝑠 because of the earth’s rotation about its axis. The centripetal acceleration of another student at Cambridge is 𝑎𝑐. What are the magnitudes of the centripetal accelerations 𝑎𝑠 and 𝑎𝑐? (Radius of earth = 6.4 x 106 m; angular velocity of earth about its axis = 7.3 x 10-5 rad s-1) Radius of the circular path followed by student in Cambridge = 6.4 × 106𝑐𝑜𝑠52° = 3.94 × 106 𝑚 𝑎𝑐 = 𝜔2𝑟 = (7.3 × 10−5)2 × 3.94 × 106 = 2.10 × 10−2 𝑚 𝑠−2 𝑎𝑠 = 𝜔2𝑟 = (7.3 × 10−5)2 × 6.4 × 106 = 3.41 × 10−2 𝑚 𝑠−2 P5. (2019 P1 Q11) D (𝑎 = 𝜔2𝑟 → 𝜔 = √ 𝑎 𝑟 = √20𝑔 7.0 = 5.3 𝑟𝑎𝑑 𝑠−1) 52° Cambridge Equator Axis of earth’s rotation

National Junior College 2022 | SH1 | H2 Physics P6. (2019 P1 Q10) B (𝑣 = 𝜔𝑟, since v is constant, 𝜔 ∝ 1 𝑅) Centripetal Force P7. (2015 P1 Q10) C (String is in tension so it must be being pulled, string assumed to have negligible weight. Note this is for forces on string not the ball. See E10 for forces on the ball)

National Junior College 2022 | SH1 | H2 Physics P8. Two identical coins are placed on a flat horizontal turn-table as shown in the figure below. Explain which coin would slip first as the angular velocity of the turn-table is increased. Centripetal force required to maintain circular motion = rm 2 Since both coins are rotating along a common radial line from the centre of the circle, their angular velocities  are constant. But Coin 2 requires a larger centripetal force to maintain its circular motion because the radius r of its circular path is larger than that of Coin 1. Since the frictional force by the turn-table on each coin is fixed, we expect that Coin 2 would slip first as the turn-table is rotated. P9. (2016 P1 Q13) D (The frictional force provides the centripetal force. Hence 𝐹 = 𝑚𝑣2 𝑟 = 𝑚(16)2 𝑟 The radius r is constant. When the maximum frictional force is halved, 𝑣2 where v is the maximum safe speed is halved too. Wet road, Fwet = 𝑚𝑉𝑛𝑒𝑤2 𝑟 =½ F 𝑚𝑣𝑛𝑒𝑤2 𝑟 = ½ 𝑚(16)2 𝑟 Hence new vnew = √162 2 = 16 √2 P10. On a normal day, the maximum friction between the wheels of a 1000 kg car and the road is 6500 N. (a) Calculate the maximum speed in km h-1 the car can go round a sharp circular bend of radius 6.0 m without skidding. The friction on the road provides the centripetal force for the car to turn. Hence maximum centripetal force = 6500 = 𝑚 𝑣2 𝑟

National Junior College 2022 | SH1 | H2 Physics Maximum speed = √6500×𝑟 𝑚 = √6500×6.0 1000 = 6.25𝑚𝑠−1 = 22.5𝑘𝑚 ℎ−1 (b) A circular bend of radius 6.0 m is indeed a very sharp bend. Usually in Singapore, most of the bends have a speed limit of 50 𝑘𝑚 ℎ−1. Calculate the minimum radius of the bend for a speed limit of 50 𝑘𝑚 ℎ−1 to be safe. 𝐹𝐶 = 𝑚 𝑣2 𝑟 𝐹𝐶 ≤ 6500 𝑚 𝑣2 𝑟 ≤ 6500 𝑟 ≥ 𝑚 𝑣2 6500 𝑟 ≥ 1000 13.892 6500 → minimum radius 29.7 m (c) Explain what will happen to your calculated value of the minimum radius if you have to take into account rainy weather. Wet weather will result in less maximum friction between the tyres and the road. As such the maximum centripetal acceleration provided will be decreased and hence the minimum radius must be larger. (d) On a racing track, cars or bicycles need to go on a speed as fast as possible. Suggest how a sharp bend can be made safer for faster speeds without increasing the friction between the tyres and the road. The bend can be a banked road and the normal contact force from the banked road will provide additional centripetal acceleration. P11. Consider a marble spinning in a horizontal circle on the inside of a cone as shown below (a) Show that the speed of the marble for circular motion at height ℎ is √𝑔ℎ. Recalling our vector diagram 𝑊 = 𝑚𝑔 = 𝑁𝑐𝑜𝑠𝜃 𝐹𝐶 = 𝑚 𝑣2 𝑟 = 𝑁𝑠𝑖𝑛𝜃 𝑡𝑎𝑛𝜃 = 𝑚𝑣2 𝑟 𝑚𝑔 = 𝑣2 𝑔𝑟 θ h normal contact force N weight W resultant force / centripetal force FC 𝜃

National Junior College 2022 | SH1 | H2 Physics 𝑡𝑎𝑛𝜃 = ℎ 𝑟 = 𝑣2 𝑔𝑟 𝑣 = √𝑔ℎ (b) Hence, state how the speed of the marble must be adjusted such that the marble is moving in a horizontal circle at a greater height. Based on the equation, a larger h will require a larger speed v. Challenging Question C1. (a) (b) Vertical components of the forces must add up to zero since this is horizontal circle 𝑇𝑢𝑝𝑝𝑒𝑟 𝑠𝑖𝑛30° = 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 sin 30° + 𝑚𝑔 35 𝑠𝑖𝑛30° = 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 sin 30° + 1.34𝑔 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 8.71 𝑁 (c) Net force is the sum of the horizontal components of the tensions 𝐹𝑐 = 𝑇𝑢𝑝𝑝𝑒𝑟 𝑐𝑜𝑠30° + 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 cos 30° = 35 𝑐𝑜𝑠30° + 8.71 cos 30° = 37.9 𝑁 (d) 𝐹𝑐 = 𝑚𝑣2 𝑟 = 37.9 1.34×𝑣2 1.70𝑐𝑜𝑠30° = 37.9 𝑣 = 6.45 𝑚 𝑠−1 weight upper tension bottom tension 60° 60°