Consider the motion of a rock tied to a string of length 0.50 m. The string is s
ID: 1429280 • Letter: C
Question
Consider the motion of a rock tied to a string of length 0.50 m. The string is spun so that the rock travels in a vertical circle as shown in the figure below. The mass of the rock is 1.5 kg, and it is twirling at constant speed with a period of 0.27 s.
(a) What is the total force on the rock directed toward the center of the circle? (Enter the magnitude only)
(b) Find the tension in the string when the rock is at the top and when it is at the bottom of the circle. (Enter magnitudes only.)
at the top at the bottomExplanation / Answer
m = 1.5 kg
mg = weight of rock = (1.5)(9.8) = 14.7 N
R = 0.50 m
T = 0.27 s means that the angular velocity of the rock = 2/0.27 = 23.3 rad/s = w
The Centripetal Force = Fc = mV²/R = m(wR)²/R = mw²R
Fc = (1.5)(23.3)²(0.50) = _________ N ANS a) {U can number crunch this, Ok?}
once U have the numerical value of Fc the Tension in the string at the:
1) Top = Fc - weight of rock = ________ N ANS b-1)
2) Bottom = Fc + weight of rock = _________ N ANS b-2)