Recent content by joe_coolish

  1. J

    Torque and angular momentum of a conical pendulum

    Ah! That works :) r(mg)sinθ = 2.73 and since the tension is θ = 0; that would be 2.73 + 0 = 2.73 :) and for Angular Momentum, it would be rxp = r(mv)sinθ? My calculation for v must be wrong. Am I using the right formula to calculate v?
  2. J

    Torque and angular momentum of a conical pendulum

    ok, this is where my brain stops working. r and F are both vectors (correct?), and I'm assuming that r is the position vector of the ball, and F is the force vector of the 2 forces acting on the ball? if so, would r= Lsinθi + Lcosθj + (?)k and F= -mgi + Tj + (?)k? How do I find the k...
  3. J

    Torque and angular momentum of a conical pendulum

    Thank you for the reply! All of the forces that are acting on the ball are from gravity and the tension of the string. As well as the centripital force pushing the ball inward. Am I missing another force? As for "playing with the numbers" I meant I didn't know what to do so I changed the...
  4. J

    Torque and angular momentum of a conical pendulum

    Homework Statement A ball (mass m = 250 g) on the end of an ideal string is moving in circular motion as a conical pendulum as in the figure. The length L of the string is 1.85 m and the angle with the vertical is 37°. What is the magnitude of the torque exerted on the ball about the...
  5. J

    2 masses around a pulley w/ friction

    Ok, I was able to get the acceloration using another thread from PF, and then from there, I calculated SQRT(2ah) and that gave me the correct velocity! a = (g(M-m))/(M+m+(0.5*Mp))
  6. J

    2 masses around a pulley w/ friction

    Homework Statement 5.Two objects, M = 15.3 kg and m = 8.69 kg are connected with an ideal string and suspended by a pulley (which rotates with no friction) in the shape of a uniform disk with radius R = 7.50 cm and mass Mp = 14.1 kg. The string causes the pulley to rotate without slipping...
  7. J

    Ball on a string with initial velocity

    Ah! I need to lay off the late nighters... Thank you! SQRT(4.81^2+(2*1.54*9.8)*(1-(SIN(RADIANS(21.7))))) = 6.493047349 It's always that dang 2!
  8. J

    Ball on a string with initial velocity

    Thank you for your reply and for the welcome! I changed the formula to Vf=SQRT(Vi^2 + lg(1 - sinθ)) which gave me this: SQRT(4.81^2+(1.54*9.8)*(1-SIN(RADIANS(21.7)))) = 5.713832509 (I like to use EXCEL for these things) Which was my original answer. Sorry, I must have copied it down...
  9. J

    Ball on a string with initial velocity

    Homework Statement A ball with mass m = 1.72 kg is attached to a peg at the origin using a length of massless string of L = 1.54 m. It is released from the top of a vertical circle with speed v0 = 4.81 m/s. What is the ball’s speed (m/s) when the string makes an angle θ = 21.7° above the...
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