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Pendulum (Conservative Forces and Potential Energy

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A mass m = 6.3 kg hangs on the end of a massless rope L = 2.13 m long. The pendulum is held horizontal and released from rest.

    I already found the speed at the bottom of the pendulum's path and that is 6.46.

    I'm stuck on:

    a. What is the magnitude of the tension in the string at the bottom of the path? [br]
    b. If the maximum tension the string can take without breaking is Tmax = 534 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.) [br]
    c.Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point).
    How fast is the mass moving at the top of its new path (directly above the peg)? [br]
    d. Using the original mass of m = 6.3 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)

    2. Relevant equations

    My book doesn't give any specific formulas for tension.

    So I have F=MA and PE=MGH


    3. The attempt at a solution
    a. I tried just MG originally and the solution checker told me that T had to be greater than MG, so I used the velocity and did .5mv^2+mg and it said it was wrong. I read some stuff online and it suggested 2MG and that was also marked wrong. I honestly have no clue whatelse to use. I've tried all of the answers both positive and negative.
    b. I didn't really know what to do for this one, I just divided the tension by gravity and the online checker said "It looks like you assumed the tension only has to hold the weight. However, the tension must be greater to accelerate the mass upward in circular motion." I really don't know where to go from there, because I think I need to use the same process as in A.
    c. I have no idea where to even start on this one. I know I'm supposed to show what work I tried, but I've just been staring at the problem. I think it will make the velocity less, but I don't know what to do.
    d. Since I didn't know how to do the problem before this, I think I need values from that, and I don't really understand tension.

    [p]

    I'm so sorry that I have so many questions, but I'm just beyond lost on this problem. I tried to go to tutoring at my school, but no one showed up to help me since it's so early in the year. I would appreciate any help what so ever.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2011 #2
    a. Well, the pendulum is moving in a circle. Ignoring the string for a moment, you want the net force to be mv^2/r, which is the formula for centripetal force. You know v, r, and m. Gravity provides a force mg downward. What should tension be so the net force is the required centripetal force upwards?

    b. When you did part (a), you got an expression for tension in terms of mass, velocity, and radius. Keep v and r plugged in but don't substitute for mass, since that's what you're looking for. Set equal to the tension and solve.

    c. This is the beauty of using energy to solve problems! You can have the weirdest, more complex forces acting and still be able to find the answer. In this case, it's pretty difficult to find what the tension actually is, so don't bother. No nonconservative forces acted, so energy (KE + PE) is conserved.

    d. This is part (a) with a little twist. Use your result from part (c).

    Good luck! If you need more help I'll post a full answer later.
     
  4. Sep 25, 2011 #3
    I'm sorry...I kind of need some more help.

    a. mv^r=121.149 and mg=61.74. I tried adding them together, but the solver said I was wrong. I'm not sure what else to do with the numbers.
     
  5. Sep 25, 2011 #4
    I got it!

    Thank you so much. I probably read your hints a hundred times, but they finally clicked!
     
  6. Sep 25, 2011 #5
    No problem- that's what this forum is for!
     
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