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Torque and Momentum Problem

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball (mass m = 250 g) on the end of an ideal string is moving in a circular motion as a conical pendulum. The length L of the string is 1.84 m and the angle with the vertical is 37 degrees.
    a) What is the magnitude of the torque (N m) exerted on the ball about the support point?
    b) What is the magnitude of the angular momentum (kg m^2/2) of the ball about the support point?

    Correct Answers: a) 2.71 b) 1.32


    2. Relevant equations

    L = m * v * r (where L = momentum)

    F centripetal = (mv^2) / r

    weight = mg

    torque = r vector x F vector

    3. The attempt at a solution

    For A I assumed the torque was 0 because there wasn't any said force. This obviously was wrong but I do not know why.

    For B I used L = m*v*r where L = momentum

    First I drew out the diagram.

    I noticed r = L * sin 37

    I noticed m*g*tan 37 = (m * v^2 /r)

    m = .250 kg
    v = (r * g * tan 37 )^(0.5) = (L*sin 37*9.8*tan 37)^(0.5) = 2.8596
    r = L * sin 37 = 1.84 * sin 37= 1.10733

    Plug and chug:

    L = .791 This was incorrect.

    Please help!
    Thanks
     
  2. jcsd
  3. Jul 11, 2012 #2
    Okay I found out that torque = L * force . The force is gravity!

    1.84 x .250 * 9.8 X sin(180 - 37) = 2.71
     
  4. Jul 11, 2012 #3
    r needs to be 1.84. The length of the string. Got it!
     
  5. Jul 11, 2012 #4
    We know that torque is calculated in the following way.

    [itex] \sum_i^n \overline{T_i} = \sum_i^n \overline{r} \times \overline{F} _i [/itex] (1)

    Also we know that there are two external forces that are applied in the ball:

    The tension of the string:

    [itex]\overline{T} = -T \widehat{x} [/itex]

    And the weight:

    [itex]\overline{P}= mg(cos \gamma \widehat{x} - sin \gamma \widehat{y})[/itex]


    Also we know that the position of the ball (b) from the support point (s) is the lenght of the string, we define it in the following way:

    [itex]\overline{r_{sb}} = L \widehat{x} [/itex]

    [itex]\overline{r_{ss}} = 0 \widehat{x}[/itex]


    Remember that the distance from the support point to the support point is zero, so with all these information we can use (1):

    [itex] \sum_i^n \overline{T_i} = \sum_i^n \overline{r} \times \overline{F} _i = \overline{r_{sb}} \times mg(cos \gamma \widehat{x} - sin \gamma \widehat{y}) + \overline{r_{ss}} \times (-T) \widehat{x} [/itex]

    Rewriting:

    [itex] \sum_i^n \overline{T_i} = L \widehat{x} \times mg cos \gamma \widehat{x} + L \widehat{x} \times - mg sin \gamma \widehat{y} + 0 \times (-T) \widehat{x} [/itex]


    [itex] \sum_i^n \overline{T_i} = L \widehat{x} \times - mg sin \gamma \widehat{y} [/itex]

    Replacing all the data that we have:

    [itex] | \sum_i^n \overline{T_i} | = (1.84 m)(9,8 \frac{m}{s^2})(0.250 kg) sin 37° \approx 2.7129 Kg \frac{m^2}{s^2}[/itex]

    For item b we know that angular momentum is defined as:

    [itex] \sum_i^n \overline{L_i} = \sum_i^n \overline{r_i} \times m \overline{v_i} [/itex] (2)

    I am not seeing how you determined that the velocity of the ball is:

    [itex]v^2 = r*g*tan 37 [/itex]

    Can you explain it?.
     
  6. Jul 11, 2012 #5
    Centripetal F = mv^2 / r.

    This force can also be written in terms of weight and the angle:

    mgtan 37

    Set them equal to each other;

    mgtan(37) = mv^2 /r

    Cancel m:

    gtan(37) = v^2 / r

    Simplify for V^2:

    rgtan(37) = v^2
     
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