Rotational Dynamics

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

prev | up | next

Discussion

Taste and compare …

  translational   rotational
1st An object at rest tends to remain at rest and an object in motion tends to continue moving with constant velocity unless compelled by a net external force to act otherwise.   An object at rest tends to remain at rest and an object in rotation tends to continue rotating with constant angular velocity unless compelled by a net external torque to act otherwise.
2nd Acceleration is directly proportional to net force and inversely proportional to mass or
∑F = m a		   Rotational acceleration is directly proportional to net torque and inversely proportional to moment of inertia or
∑τ = I α
3rd For every action there is an equal and opposite reaction. (Here action and reaction refer to forces.)   For every action there is an equal and opposite reaction. (Here action and reaction refer to torques.)

 

Translational and Rotational Quantities Compared
  translational connections rotational
cause of acceleration
∑ F

  
τ =
r × F
∑ τ

  
resistance to acceleration
m

  
I =
∑ ri2mi = ∫ r2 dm
I

  
newton's second law
∑ F =
m a

    
∑ τ =
I α

Start with the equation for Newton's second law. Cross radius into both sides and then play around with the algebra. (This requires some unusual identities that I've seen before, but never really learned.)

∑F =  m a
∑r × F =  m r × a = m r × (α × r)
∑r × F =   m α(r·r) − r(r·α) = m α r2 − 0 = mr2 α
∑τ =  I α

In the end, we get an analogous formula for Newton's second law of rotation with two new quantities: torque, the rotational equivalent to force

τ = r × F

and moment of inertia, the rotational equivalent to mass

I = ∑ ri2 mi =  ⌠
⌡		  r2 dm

More …

Summary

Problems

practice

  1. A kind of Atwood's machine is built from two cylinders of mass m1 and m2; a cylindrical pulley of mass m3 and radius r; a light, frictionless axle; and a piece of light, unstretchable string. The heavier mass m1 is held above the ground a height h and then relased from rest.
    1. Draw a free body diagram showing all the forces acting on …
      1. the heavier mass
      2. the lighter mass
      3. the pulley
    2. Write the equation stating Newton's second law of translational motion for …
      1. the heavier mass
      2. the lighter mass
      and rotational motion for …
      1. the pulley
    3. Determine the translational acceleration of …
      1. the heavier mass
      2. the lighter mass
      and the rotational acceleration of …
      1. the pulley
    4. Determine the tension in the side of the string connected to
      1. the heavier mass
      2. the lighter mass
      and the upward force of the axle on …
      1. the pulley
    5. Lastly, determine …
      1. the time it takes for the heavier mass to reach the ground
      2. its speed on impact
      3. the rotational speed of the pulley at this time

    This might seem like a big problem, but it's actually just a bunch of small ones. Since problems in rotational dynamics tend to get complicated very quickly, it seems like a good way to introduce this topic.

    1. Answer it.
      1. Answer it.
      2. Answer it.
      3. Answer it.
    2. Answer it.
      1. Answer it.
      2. Answer it.
      3. Answer it.
    3. Answer it.
      1. Answer it.
      2. Answer it.
      3. Answer it.
    4. Answer it.
      1. Answer it.
      2. Answer it.
      3. Answer it.
    5. Answer it.
      1. Answer it.
      2. Answer it.
      3. Answer it.
  2. A roll of toilet paper is held by the first piece and allowed to unfurl as shown in the diagram to the right. The roll has an outer radius R = 6.0 cm, an inner radius r = 1.8 cm, a mass m = 200 g, and falls a distance s = 3.0 m. Assuming the outer diameter of the roll does not change significantly during the fall, determine …
    1. the tension in the sheets
    2. the translational acceleration of the roll
    3. the angular acceleration of the roll
    4. the final translational speed of roll
    5. the final angular speed of the roll

    There are two general approaches to this problem: using Newton's second law (see below) and conservation of energy (see the section on rotational energy).

    1. Apply Newton's second law of motion in both its translational and rotational forms. (Let down be the positive direction.)
             
      translational rotational
      ∑F =  m a ∑τ =  I α
      mg − T =  ma RT =  1  m(R2 + r2)  a  
      2 R  
                   
      Now work the magic of algebra. Begin by rewriting the rotational equation a bit then substitute from the translational side and solve for tension. This is not as easy to do as it is to say, however. Those of you not comfortable with the algebra of pure symbols may want to substitute numbers in now, simplify a bit, and then solve. Whatever gets you the right answer is probably a right method.
         
      2R2T =  (R2 + r2)ma = (R2 + r2)(mg − T)
      2R2T + (R2 + r2)T =  (R2 + r2)mg
      T =  (R2 + r2)mg  
      3R2 + r2
      T =  [(0.060 m)2 + (0.018 m)2](0.200 kg)(9.8 m/s2)
      3(0.060 m)2 + (0.018 m)2
      T =  0.691 N
         
    2. Jump back to the translational statement of Newton's second law then jump through the usual hoops: algebra, numbers, answer. The answer should be less than the acceleration due to gravity (since tension is dragging upward against weight).
         
      mg − T =  ma
      a =  g −  T  = 9.8 m/s2 −  0.691 N
      m 0.200 kg
      a =  6.34 m/s2
         
    3. At last, an easy problem. Just be sure to use the outer radius of the roll, not the inner.
               
      α =  a  =  6.34 m/s2  = 106 rad/s2 = 16.8 rev/s2
      R 0.060 m
               
    4. Another easy problem. At least, you should find it easy if you've gotten this far through this book.
         
      v2 =  v02 + 2aΔs
      v =  √(2 × 6.34 m/s2 × 3.0 m)
      v =  6.17 m/s
         
    5. One more easy problem. There are several ways to solve this part. Whatever method you chose, use the correct radius.
               
      ω =  v  =  6.17 m/s  = 103 rad/s = 16.4 rev/s
      R 0.060 m
               
  3. The top shown below consists of a cylindrical spindle of negligible mass attached to a conical base of mass m = 0.50 kg. The radius of the spindle is r = 1.2 cm and the radius of the cone is R = 10 cm. A string is wound around the spindle. The top is thrown forward with an initial speed of v0 = 10 m/s while at the same time the string is yanked backward. The top moves forward a distance s = 2.5 m, then stops and spins in place.
    [magnify]
     
    Using rotational dynamics (and kinematics) determine …
    1. the moment of inertia I of the top (essentially, the moment of inertia of a cone)
    2. the tension T in the string
    3. the final angular velocity ω of the top
    4. the length l of string wound around the spindle

    Solutions …

    1. We have a very nice formula for the moment of inertia of a cone. It would be a shame to ruin it with messy numbers, but compute we must.
               
      I =  3  MR2 =  3  (0.50 kg)(0.10 m)2= 0.0015 kgm2
      10 10
               
    2. Use the second law of translational motion. The net force on the top comes from the tension of the string (since weight and normal force cancel and friction is assumed negligible). Replace acceleration with a suitably rearranged version of one of the translational equations of motion. Substitute and solve.
                   
      ∑F =  ma   v2 = v02 + 2aΔs   a =  v2
      2Δs
      T =  mv2  =  (0.50 kg)(10 m/s)2  =  10 N
      2Δs 2(2.5 m)
                   
    3. Not only does the tension supply the net external force needed to stop the forward motion of the top, it also applies the net external torque needed to get the top spinning. We need to explore both realms of Newton's second law to solve this part of the problem.
             
      translational rotational
      ∑F =  ma ∑τ =  Iα
      T =  ma rT =  Iα
             
      On the translational side, replace acceleration with an equation of motion that can be used to find time.On the rotational side, replace angular acceleration with an equation of motion that uses time.
                 
      translational rotational
      v =  v0 + aΔt ω =  ω0 + αΔt
      a =  v0   α =  ω        
      Δt Δt
      T = m  v0   rT = ⎛
      ⎝      		 3  mR2 ⎞⎛
      ⎠⎝      		 ω ⎞
      ⎠
      Δt 10 Δt
                       
      Now, combine the two formulas by substituting T from the translational equation into T in the rotational equation, then watch stuff drop out. Do not cancel the radii, however. One is the radius of the spindle (r) and the other is the radius of the base (R).
                             
      r  ⎛
      ⎝      		 m  v0 ⎞
      ⎠      		  =  ⎛
      ⎝      		 3  mR2 ⎞⎛
      ⎠⎝      		 ω ⎞
      ⎠
      Δt 10 Δt
      rv0  =    3  R2ω      
      10
                             
      Solve for ω, input numbers, and compute the answer. (Pay attention to the units.)
               
      ω =  10rv0  =  10(0.012 m)(10 m/s)  = 40 rad/s = 6.37 rotations per second
      3R2 3(0.10 m)2
               
    4. Find the time first. How long was the string in contact with the top? Use translational equations for this.
               
      s =  v + v0   Δt
      2
      Δt =  2s  =  2(2.5 m)  = 0.50 s
      v + v0 10 m/s
               
      Pop this number into its rotational equivalent.
           
      θ =  ω + ω0   Δt
      2
      θ =  0.50 rad/s  0.50 s = 10 rad
      2
           
      Multiply by the radius of the spindle to determine the length of the string.
       
      l = rθ = (0.012 m)(10 rad) = 0.12 m = 12 cm
       
  4. Write something different.
    • Answer it.

conceptual

  1. Describe a test involving rotational principles that can be used to distinguish raw eggs from hard boiled eggs.
  2. Some airplanes have only one propeller, but all helicopters have two rotors. (A helicopter with only one rotor is unflyable.)
    1. Why must helicopters always have two rotors?
    2. How does a single propeller aircraft manage to avoid the problems encountered by a single rotor helicopter?

worksheets

  1. The Physics Teacher has published several articles containing free body diagram worksheets. They are available free to members of the American Association of Physics Teachers (AAPT). Everyone else has to pay.
    1. Free-body diagrams revisited — II. James E. Court. The Physics Teacher. Vol. 37, No. 8 (November 1999): 490-495. Note: pages 494-495 are relevant to this topic.

Resources

prev | up | next

Another quality webpage by

Glenn Elert		 eglobe logo home | contact

bent | chaos | eworld | facts | physics