Transcription of Circular Motion Problems ANSWERS
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Circular Motion Problems ANSWERS
Ftension Fgravity= N Circular Motion Problems ANSWERS 1. An g cork is swung in a horizontal circle with a radius of 35 cm. It makes 30 revolutions in 12 seconds. What is the tension in the string? (Assume the string is nearly horizontal) T=time/revolutions= s Period is the time per revolution F=ma Write down N2L Ftension = mv2/r Tension provides net force, acceleration is centripetal Ftension=m(4 2r/T2) Speed equals circumference divided by period Ftension= N Substitute values and calculate 2. A 15 g stopper is swung in a horizontal circle with a radius of meters. The tension in the string is Newtons. Find the speed of the stopper and determine how long it takes to complete 30 revolutions. (Assume the string is very nearly horizontal). F=ma Write down N2L Ftension = mv2/r Tension provides net force, acceleration is centripetal v=sqrt(Ftension r / m) = m/s Solve for v and calculate v=2 r/T T=2 r/v = s Speed equals circumference divided by period t = 30 T = s Time = # rev x seconds/rev 3.
Dec 11, 2012 · The visible mass in that galaxy is 1.8x1041 kg. What fraction of the galaxy’s mass is dark matter? F=ma N2L F gravity =mv 2/r Gravity provides the centripetal acceleration GM galaxy m star /r 2=m star v 2/r Star is the mass moving in a circle M galaxy =v 2r/G = 8.6x1041 kg Solving for mass of the galaxy* ( Initially I goofed up my
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