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Homework Help: Some problems with rotational motion

  1. Nov 13, 2005 #1
    1) A .65 kg on the end of thin light rod is rotated in a horizontial circle of a radius 1.2 m. (a) Calculate the moment of inertia of the ball about the center of the circle, and (b) the torgue needed to keep the ball rotating at constant angular velocity if air resistence exerts a force of .020 N on the ball. (ignore rod's moment of I and air resistence).

    So... I did find (a) successfully: I = mr^2 = .94 kg-m^2

    And here were my thoughs for part (b): T = F(perpindic)r
    T = I*a(angular) Since acceleration is constant is it not zero? correct answer: .0240 mN

    2)To get a flat uniform cylindrical satellite spinning at the correct rate, engineers fire four tangential rockets. If the satellite has a mass of 3600 kg and a radius of 4 m, what is the require steady force of each rockets if the satellite is to reach 32 rpm in 5 minutes?

    This one really got me thinking. So first converted all I needed to. I was give:
    m = 3600 kg, r = 4 m, w = 3.349 rads/2, = 300 s.

    I know I have to use the F = ma, and I have the m so all I need to do is some how get the a, and that is where I am having trouble. Correct answer: 20 N

    3) A bowling ball of mass 7.3 kg and radius 9 cm = .09 m, rolls without slipping down a lane at 3.3 m/s. Calculate it's total kinetic energy.

    What I did was use the equation: (1/2)Iw^2

    First I found I by I=mr^2
    I = .059

    then I found w by w = v/r = 3.3/.09
    w = 36.6

    so .5(.059)(36.6^2)
    = 39.5 J and the correct answer is 56 J What did I do wrong?
  2. jcsd
  3. Nov 13, 2005 #2


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    for problem 3,

    one needs to find the translational and rotational KE's.

    Obviously, KE (trans) = 1/2 mv2.

    Rotational KE = 1/2 I [itex]\omega[/itex]2.

    Make sure one has proper moment of inertia - in this case for a sphere I = 2/5 mR2, and when one multiplies R2 by [itex]\omega[/itex]2, the product is just v2.

    For problem 2, try torque T = I[itex]\,\alpha[/itex], and [itex]\omega[/itex] = [itex]\omega_o[/itex] + [itex]\alpha\,[/itex]t, where [itex]\alpha[/itex] is angular acceleration and [itex]\omega[/itex] is angular velocity, and [itex]\omega_o[/itex] is the initial angular velocity.

    For problem 1,
    To keep the ball moving, the applied torque must equal the opposing torque, which you have stated as T = F(perpindic)r. Simply subsitute in the values given. I think the units given with the correct answer are not correct, but should be N, not mN.
  4. Nov 14, 2005 #3
    Problem three:
    .5mv^2 = (.5)(7.3 kg)(3.3 m/s)^2 = 39.7 J

    .5Iw^2 = finding I = 2/5(7.3 kg)(.09 m)^2 = .025

    so... .5(.025)(w) and w = v/r = (3.3 m/s)/(.09) = 37
    so (.5)(.025)(36.6) = 16.7

    adding them together is 16.7 J + 39.7 J = 56.4 = 56 J. Would that be correct?

    #2 I used w = w(i) + a(ang)t
    3.349 rad/s = 0 + a(300 s )
    ang. acc = .011 m/s^2

    then I found I, which fora cylinder is 1/2mr^2
    = .5(3600 kg)(4 m)^2 ---- I = 28800

    then I substitute in T = (28800)(.011 m/s^2) = 316... still not the right answer, why is that?
    #1, T = F(perpindic)r = (.020 N)(1.2 m) = .024 Nm, that is correct! At least I got two of them understood. Thanks. Any idea on # 2? Also, why are we only considering the force the air resistance is creating for #1 and now the mass?
  5. Nov 14, 2005 #4


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    Try this on #2,

    One is trying to calculate the Torque provided by 4 rocket motors (jets) at radius 'r', tangential to the circumference of the satellite (cylinder).

    So torque T = 4 F x r, which is the total force at a moment arm of length r. Also the torque = Moment of Inertia x angular acceleration, which you have calculated correctly.

    Equate 4 F r = I [itex]\alpha[/itex] and solve for F.

    On problem #1, I misread your units. I was thinking mN meant milliNewtons, rather than N-m (Newton-meters), which is the correct unit for torque.
  6. Nov 14, 2005 #5
    Okay I shall try again.
    4 F r = I [itex]\alpha[/itex]

    4(F)(4 m) = (28800)(.011 m/s^2)
    = 19.8 = 20 N, did I do the equation correctly?
  7. Nov 14, 2005 #6


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    That seems to be correct.
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