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 1