Rotational Dynamics

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.

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1. Apply Newton's second law of motion in both its translational and rotational forms. (Let down be the positive direction.)

   translational	rotational
   ∑F =  ma     mg − T =  ma	∑τ =  Iα     RT =  ½m(R2 + r2)  a 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
   2R2T =	(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
   		m

   a = 9.8 m/s2 −		0.691 N
   		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 R
   α =	6.34 m/s2
   	0.060 m
   α = 106 rad/s2
   α = 16.8 rev/s2

4. Another easy problem. At least, you should find it easy if you've gotten this far through this book.

   v2 =	v02 + 2a∆s
   0v =	√(2 × 6.34 m/s2 × 3.0 m)
   0v = 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 R
   ω =	6.17 m/s 0.060 m
   ω = 103 rad/s
   ω = 16.4 rev/s

1. We already have a 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
   	10
   I =	3	(0.50 kg)(0.10 m)2
   	10
   I = 0.0015 kg m2

2. Use the second equation 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 = T = ma	&	v2 = v02 + 2a∆s

   T =	mv2 2∆s
   T =	(0.50 kg)(10 m/s)2
   	2(2.5 m)
   T = 10 N

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 T =  ma	∑τ =  Iα 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 + a∆t     a =  v0 ∆t T =  m  v0 ∆t	ω =  ω0 + α∆t     α =  ω ∆t rT =  ⎛ ⎜ ⎝ 3  mR2 ⎞⎛ ⎟⎜ ⎠⎝ ω ⎞ ⎟ ⎠ 10 ∆t

   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	⎞ ⎟ ⎠
   			∆t				10			∆t

   rv0 =	3	R2ω
   	10

   Solve for ω, input numbers, and compute the answer. (Pay attention to the units.)

   ω =	10rv0 3R2
   ω =	10(0.012 m)(10 m/s)
   	3(0.10 m)2
   ω = 40 rad/s
   ω = 6.37 rotations per second

4. Find the time first. How long was the string in contact with the top? Use translational equations for this.

   s =	v + v0   ∆t 2
   ∆t =	2s v + v0
   ∆t =	2(2.5 m)
   	0 m/s + 10 m/s
   ∆t =	0.50 s

   Pop this number into the rotational equivalent of the previous equation.

   θ =	ω + ω0	∆t
   	2
   θ =	0.50 rad/s	0.50 s
   	2
   θ =	10 rad

   Multiply by the radius of the spindle to determine the length of the string.

   ℓ = rθ ℓ = (0.012 m)(10 rad) ℓ = 0.12 m = 12 cm

Answer it.