Lecture 11Dymanics of circular motion(1)
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ECE 105October 3, 2022Lecture #11Dynamics of Circular Motion
Circular motion: ππͺ= ?????=??πΉππͺ=???πΉABRiο΅fο΅O??π?
Actual and fictitious forces π?π??π?π????Note, that even in the case of an object undergoing uniform circular motion, there is a change in the direction of the vector velocity. Therefore the object is moving in an accelerated frame of reference accelerating with centripetal acceleration, ac, toward the center of the circular trajectory.
Circular motion: plane parallel to ground A metal ball with mass M = 0.2 kg attached to a string is twirled around horizontally a at a rate of 2 revolutions / sec. The length of the string is R = 25cm.Q. What is the tension on the string?Given:β’M = 0.2 kgβ’R = 0.25 mβ’v = 2 rev / sFind:β’T= ? π» = ?. ? Γ(?. ??)??. ??= ?. ??πΉπ»ΰ·π=?????=??πΉπ»=???πΉπ» = ????for1rev/sec=?π
πΉ?(Ξ€??)? = ??π
πΉ (Ξ€? ?)? = ? Γ ?π
Γ ?. ?? = ?. ????For N rev/sec
Circular motion: plane perpendicular to ground ΰ·π?=??????π?π½ = ?????=??π?π½ΰ·ππͺ=??πͺπ»β?????π½=??πͺ=???πΉπ»=?????πΉ+???π½π»???=????????πΉβ?π»??????=???????????πΉ+?Consider an intermediate point of the trajectory, where the string is at angle ΞΈrelative to the vertical line. We can evaluate the tangential acceleration, atand the tension on the string at that point as functions of ΞΈ. Now, consider the tension on the string at the top and bottom positions:
Circular motion: Problems In the movie βTop gunβ ,Tom Cruise is a Navy pilot with a mass M = 75 kg. During a training exercise he performs a dive with a speed v = 25 m/s. Assume that the trajectory of the plane is a circle with a radius R = 500 m.Q1. How large is the normal force exerted by the seat onto the pilot at the lowest point of the dive?Given:β’M = 75 kgβ’v = 25 m/sβ’R = 500 mFind:β’FN = ? R+y?? β π?= β???π?= ?? + ???π?= ?(? + ??)??=??πΉπ?=?(?+??πΉ)π?=???.?+??????=???.??π??π???
Circular motion: A child with mass M = 40 kg is riding a roller-coaster in the form of a spiral with circular cross-section and radius R = 70 m. Q. What is the minimal linear velocity of the roller-coaster at which the child is about to fall off the chair at the highest point of the circle?Given:β’M = 40 kgβ’R = 70 mFind:β’v= ? +y??π????π + πΉπ= ?ππ?π=?ππ=?π£2?ππ=π£2?π£ =π? =9.8 Γ 70 = 26ππ The condition for the child to about to fall off the seat is: FN= 0
Circular motion: A Ferris wheel rotates at 1 revolution per 30s and has a radius of 5m.Q1. What is the force of the seat on the child at the top and bottom points of the ride?Given:β’M = 45 kg β’R = 5 mβ’v = 1 rev/30 s Find:β’FN = ? ΰ·πΉ=?ππ=?π£2??π β πΉπ= β?π£2?πΉπ= ?π +π£2?π£ =??=2π?30= 1.05Ξ€ππ πΉπ= 45 Γ9.8 +(1.05)25= 451?????+yBottompointToppoint?π β πΉπ= ?π£2?πΉπ= ?π βπ£2?π£ =??=2π?30= 1.05Ξ€ππ πΉπ= 45 Γ9.8 β(1.05)25= 431?
Circular motion: A car rounds a curve on a flat road with radius R = 90 m, with speed of 100 km/h.Q. What is the minimal static friction coefficient necessary to keep the car from sliding sideways?Given:β’R = 90 m β’v = 100 km/hFind:β’ΞΌs= ? πΉΰ·π?= ???πΉ??= ???πΉ(?)??= π?π?(?)???πΉ= π?π?(?)??ΰ·π?= ?π?β ?? = ? (?)π??? = ???πΉπ?=???πΉπ?=ΰ΅????.???.?Γ??=?.??Substitute eq. (2) in eq. (1) Combine eq. (3) and eq. (4) and remove FN
Circular motion: A car rounds a banked curve. The radius of curvature of the road is R, the banking angle is ΞΈand the coefficient of static friction is ΞΌs.Q. What is the range of speeds the car could have without slipping up or down the road?Given:β’Rβ’ΞΈβ’ΞΌsFind:β’vmin = ?β’vmax = ?YX?π If the car moves on the circular path at low speed, it will tend to slide downhill. Therefore, friction will point uphill.ΰ·π?= ???πΉπ??π’?π½ β ??π??π½= ???π??πΉ(?)??= π?π?(?)??π?=πΉπ??(?π’?π½βπ?π??π½)(?)Substitute eq. (2) in eq. (1) and express vmin: ΰ· π?= ?π?π??π½ + ???π’?π½ β ?? = ? (?)??= π?π?(?)π?π??π½ + π?π??π’?π½ β ?? = ?π?=??π??π½ + π??π’?π½(?)??π?=?πΉ?π?π½βπ??+π??π?π½Substitute eq. (5) in eq. (4)Substitute eq. (6) in eq. (3)?π
Circular motion: A car rounds a banked curve. The radius of curvature of the road is R, the banking angle is ΞΈand the coefficient of static friction is ΞΌs.Q. What is the range of speeds the car could have without slipping up or down the road?Given:β’Rβ’ΞΈβ’ΞΌsFind:β’vmin = ?β’vmax = ?YX?π If the car moves on the circular path at very high speed, it will tend to slide uphill. Therefore, friction will point downhill.ΰ·π?= ???πΉπ??π’?π½ + ??π??π½= ??????πΉ(?)??= π?π?(?)????=πΉπ??(?π’?π½+π?π??π½)(?)Substitute eq. (2) in eq. (1) and express vmax: ΰ· π?= ?π?π??π½ β ???π’?π½ β ?? = ? (?)??= π?π?(?)π?π??π½ β π?π??π’?π½ β ?? = ?π?=??π??π½ β π??π’?π½(?)????=?πΉ?π?π½+π??βπ??π?π½Substitute eq. (5) in eq. (4)Substitute eq. (6) in eq. (3)?π