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# uniform circular motion answer key

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When a body moves with constant speed in a circle of radius \$r\$, the motion is uniform circular motion. To explore this idea further, this week Shini sits down with us to discuss centripetal force, centrifugal force, and a few other bits, This worksheet guides students through the process of solving circular motion problems.

Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? I mean, it's a thing, it's just not real. What about the instantaneous acceleration for non-uniform circular motion? flashcard set{{course.flashcardSetCoun > 1 ? Determine the magnitude of the acceleration of the car.a = v2/r T = 2πr/v.....r = Tv/2πcombine...a = v2/(Tv/2π)= v/(T/2π)a = (60)/(50/6.28) = 7.5 m/s2, 2. THERMODYNAMICS

Introduction: The acceleration toward the center that keeps objects in uniform circular motion (circular motion at a constant speed) is called centripetal acceleration.

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1. In a uniform circular motion map, what is always true? a speed v. When you run the simulation a vector diagram will appear showing the vectors. Please take your time and answer it completely.Measurement devices needed:A long measuring tapeA stopwatchStep 1: Use the measuring tape to determine the radius (r) of the path of thecar on the circular racetrack. These problems address horizontal, vertical, banked, and incorporate friction into uniform circular motion. Suppose a particle moves in a circle of radius \$r\$ with constant speed as in Figure 4. As a member, you'll also get unlimited access to over 83,000 lessons in math, Services, Newton's First Law of Motion: Examples of the Effect of Force on Motion, Quiz & Worksheet - Uniform Circular Motion, Uniform Circular Motion: Definition & Mathematics, {{courseNav.course.mDynamicIntFields.lessonCount}}, What is Kinematics?

Fig.1 above refer to a point moving along a circular path. The worksheet covers uniform circular motion and can be used in any level of a physics class. These problems address horizontal, vertical, banked, and incorporate friction into uniform circular motion. This occurs at its maximum level when the radius is 10% too low and the time measurement is 10% too high:Example:If the actual radius is 10.0 m and the radius measurement is 9.0 mIf the actual time is 10.0 s and the time measurement is 11.0 sActual acceleration = a = 4π2(10.0)/(10.0)2 = 0.40π2 m/s2Measured acceleration = am = 4π2(9.0)/(11.0)2 = 0.30π2 m/s2% Error = [(Measured – Actual)/Actual]100%% Error = [(0.30π2 – 0.40π2)/0.40π2]100% = -25 %, When the other extremes occur, the % error is in between the calculations shownabove.Example:If the actual radius is 10.0 m and the radius measurement is 11.0 mIf the actual time is 10.0 s and the time measurement is 11.0 sActual acceleration = a = 4π2(10.0)/(10.0)2 = 0.40π2 m/s2Measured acceleration = am = 4π2(11.0)/(11.0)2 = 0.36π2 m/s2% Error = [(Measured – Actual)/Actual]100% % Error = [(0.36π2 – 0.40π2)/0.40 2]100% = -10 %Example:If the actual radius is 10.0 m and the radius measurement is 9.0 mIf the actual time is 10.0 s and the time measurement is 9.0 sActual acceleration = a = 4π2(10.0)/(10.0)2 = 0.40π2 m/s2Measured acceleration = am = 4π2(11.0)/(11.0)2 = 0.44π2 m/s2% Error = [(Measured – Actual)/Actual]100%% Error = [(0.44π2 – 0.40π2)/0.40π2]100% = +10 %, When the measurements are not at the extremes, the % error again will fall inbetween its maximum positive and maximum negative values.Example:If the actual radius is 10.0 m and the radius measurement is 10.5 mIf the actual time is 10.0 s and the time measurement is 9.5 sActual acceleration = a = 4π2(10.0)/(10.0)2 = 0.40π2 m/s2Measured acceleration = am = 4π2(10.5)/(9.5)2 = 0.47π2 m/s2% Error = [(Measured – Actual)/Actual]100%% Error = [(0.47π2 – 0.40π2)/0.40π 2]100% = +18 %, 6.