HCI 06 Circular Motion Solutions (Lecture Examples)

Example 1 A Blu-ray player is spinning a standard Blu-ray Disc (120 mm diameter) at 900 revolutions per minute (RPM). Determine i) the frequency of the rotation (in revolutions per second) ii) the period of the rotation iii) the angular speed i) f = 11 900 15 rev s60 −==T ii) T = 11 0.0667 s15==f iii) 12 2 (15) 94.2 rad sππ −= = =ωf Example 2 Dishes are placed on a Lazy Susan at a reunion dinner during Chinese New Year. When the Lazy Susan is spun around to serve the guests, what can be said about the dishes in the inner circle compared to those in the outer circle in terms of their speed and angular velocity? Angular velocity The angular velocity ω is the same for all dishes. This should be clear if you realize that every dish undergoes the same angular displacement in the same amount of time. You may also notice that all dishes complete the same number of revolutions in the same duration of time. With the same T and f, they also have the same angular velocity 2 2 fT == . Speed The dishes in the inner circle have a lower speed compared to the dishes in the outer circle. As v = rω, the smaller the radius, the lower the tangential velocity.

Example 3 - Conical pendulum In a conical pendulum system, a small pendulum bob of mass 0.50 kg is rotating in a fixed horizontal plane. The string is 30 cm long and makes an angle of 15 to the vertical. Calculate the (a) tension in the string; (b) linear speed of the bob; (c) period of rotation of the bob. a) Vertically, object is stationary Net force in the vertical direction = 0 N cos os15 (0.50)(9.81) 5.1 N T θ mg Tc T = = = b) Net force in the horizontal direction provides the required centripetal force 2 sin = mvT θ r 2 1sin15 (0.30sin15 )(5.078)(sin15 )sin15 0.45 m s 0.50 −  =  = = =mv rTTv rm c) Using π π π = =  = = =2 2 2 (0.30sin15 )( ) 1.1 s 0.4498 rvr ω r T Tv Example 4 Explain, with the aid of a diagram, why the mass at the end of a light inelastic string cannot be whirled in uniform circular motion in such a way that the string is horizontal. When string is horizontal, θ = 90° The tension in the string T will no longer have a vertical component. There is no longer an upward vertical component to balance the weight of the bob. Altnernatively, T cos θ = mg As θ 90 , cos 0θ→  → Thus →T So infinitely large tension force is required to maintain vertical equilibrium if the string were horizontal. 15 15 T mg Tsin15° Tcos15° T θ mg

Example 5 - Roller coaster Some roller coasters have several loops along the track. The picture on the right illustrates such a coaster executing a loop-the-loop. In the question below, assume that the total mass of one car plus its passengers is 170 kg and that friction can be ignored. (a) A passenger car for a roller coaster enters a loop of radius 19 m at position 1, as indicated in the diagram, with a speed of 33 m s-1. Determine the normal contact force the track exerts on the car at i) the bottom (position 1) and ii) the top (position 2) of the loop. (b) Find the minimum speed at which the passenger car must travel while it is at the top of the loop, in order to clear the loop safely. (a)(i) At position 1, required centripetal force is upward. Applying N2L on car: 2 22 4 () 33(170)(9.81 ) 1.1 10 N19 netF ma mvN mg r mvN mg r = −= = + = + =  (a)(ii) Firstly, we use PCOE to find the speed at position 2. 22 2 22 2 1 2 loss in KE gain in GPE 11 (2 )22 1(33 ) (9.81)(2 19)2 18.5 m s mu mv mg r v v − = −= − =  = At the top, the net force is downward (in the centripetal direction). Applying N2L to the car: 2 2 2 2 32 () 18.53(170)( 9.18) 1.40 10 N19 netF ma mvN mg r mvN mg r = += = − = − =  r 1 2 N mg N mg

b) To just clear the loop safely, the car must not lose contact with track (i.e. N > 0). Apply N2L to car: 2 min 2 min 1 min () 0 (19)(9.81) 13.7 m s netF ma mvN mg r mvmg r v rg − = += += = = = Example 6 – String and bob A stone of mass 800 g is tied to one end of a string and is whirled in a vertical circle. The string is inextensible and of length 1.2 m. The stone has a certain speed vA at the lowest point, as shown below. (i) Determine the minimum speed the stone must have at the top of the circular motion if the string is to be taut at that instant. (ii) Hence, show that the stone can complete a vertical circular motion if vA = 8.0 m s-1. (i) At the top of the circle, According to Newton’s 2nd Law, T + mg = 2 mv r (The slower the speed at the top, the lower the tension T, when speed is a minimum, T = 0.) 0 + mg = 2 minmv r 1 min (1.2)(9.81) 3.43 m sv rg −  = = = (ii) Firstly, we use PCOE to calculate the speed at which the stone arrives at the top. 2211 22 22 1 (2 ) loss in KE gain in GPE 1(8.0 ) (9.81)(2 1.2)2 4.11 m s AB B B mv mv mg r v v − −= = − =  = Since the stone arrives at the top at a speed higher than 3.43 m s-1, it is able to complete the vertical circular motion. 1.2 m vA T mg N mg

Example 7 - Car going around a bend (a) A bend in the road has a 50 m radius of curvature. A car of mass 600 kg takes the bend at 45 km h−1. i) What is the centripetal acceleration of the car? ii) What is the centripetal force experienced by the car and what provides it? iii) What will happen to the car if driver decides to take the bend at 60 km h-1 instead? (Given that the maximum friction between the tyres and road surface is 3000 N.) (b) When the car negotiates a corner on horizontal ground, the frictional force between the tyres and the ground is the only force providing the centripetal force. As there is a limit to this frictional force for a particular road surface, there is a maximum speed which the car can make the turn safely, above which skidding will occur. Hence, some corners (especially at race -tracks) have raised embankments to increase the maximum speed at which a vehicle can take the corner than if on a level road. It does s o by making the normal contact force contribute a component to the centripetal force. For an embankment inclined at 20 to the horizontal, find the speed at which the normal contact force is able to completely provide the centripetal force (i.e. no frictional force required). (i) 3 2 2 2 45 10()60 60 3.1 m s50 c va r −  = = = (ii) 2 (600)(3.125) 1875 N 1900 N (2 s.f)cc mvF ma r= = = = = The centripetal force is provided by the friction exerted by the road surface on the car tyres. (iii) If car travels at 60 km h-1 Centripetal force required , 3 2 2 60 10(600)( )60 60 3300 N (2 s.f)50 c mvF r  = = = As the maximum frictional force between road surface and tyres is 3000 N, the frictional force is insufficient to provide the required centripetal force of 3300 N, the car will skid off the road from the circular curvature path. Centre of circular path 20 Normal contact force Frictional force Weight Normal contact force W 7 (b) Vertically, car is in translational equilibrium Vertical component of normal contact force = Weight cosN θ mg= (600)(9.81) 6264 Ncos cos(20) mgN θ= = = Horizontally, according to Newton’s 2nd Law 2 Horizontal component of normal contact force mv r= −=  = = = 2 1sin 50(6264)sin(20)sin 13 m s 600 mv rN θN θv rm The speed at which normal contact force is able to completely provide