Question

3. As shown on the right, a small 150. gram mass is attached to the end of a low mass string so that the center of the mass is 120. cm below the attachment point. The mass is pushed to that it moves in a horizontal circle forming a "conical pendulum" witha half-cone angle of 32.0°. (10 points) 320 a. What is the tension in the string? (5 points) m # 1 50. gramis answer: T- (5 points) b. How long does it take the mass to complete one revolution?

Answer 1

By Free Body Diagram T'Cos theta = mg-(1) T'Sin theta = mv^2/r-(2) T'rightarrow tension By eqn (i)-T'= mg/Cos 32 degree = 0.15 times 9.8/0.85 T = 1.73 N tan theta = v^2/rg v = squareroot rg tan theta 2pi r/T = squareroot rg tan theta T = 2pi r/squareroot rg tan theta T = 2pi squareroot r/g tan theta \r\n Time-Period = T = 2pi squareroot r Cos theta/g sin theta Sin theta = (r/l) T = 2pi squareroot r Cos theta times l/g times r T = 2pi squareroot l Cos theta/g T = 2pi squareroot 1.2 times Cos 32 degree/9.8 T = 2.03 sec For Query Do comment Thank you!!