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Because I am studying on my own, I don't have anyone to talk to about this when I don't understand, and I was wondering if someone could help me with a concept in rotational kinematics:

At the start of your throw of a $2.7\:\mathrm{kg}$ bowling ball, your arm is straight behind you and horizontal. Determine the rotational acceleration of your arm if the muscle is replaced. Your arm is $0.64\:\mathrm{m}$ long, has a rotational inertia of $0.48\:\mathrm{kg\:m^2}$, and has a mass of $3.5\:\mathrm{kg}$ with its center of mass $0.28\:\mathrm{m}$ from your shoulder joint.

I'm not interested in the answer, but I am interested in learning how I should treat the arm-bowling ball system. Do I treat the arm as its own rotating object with its own moment of inertia and the bowling ball as its own object with the rotational inertia of a hoop with rotational axis through its center ($I=MR^2$)? Or do I simply add the torques of each and treat this value as the arm's total torque?

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  • $\begingroup$ Welcome to Physics StackExchange. If this is a homework problem include the homework-and-exercises tag. Your question seems a little broad, especially the part about "how to treat things when solving these types of problems"... Usually the more precise questions are more easily answered in this forum. Hopefully someone has time to understand and address your question, unfortunately that somebody is not me. Good luck. $\endgroup$
    – hft
    Apr 6, 2015 at 2:21
  • $\begingroup$ The way to understand this is the same as the way to understand any other basic dynamics problem: draw the free-body diagram(s), label the forces, apply $\sum \vec{F}_i = m\vec{a}$ and $\sum \tau_i = I \alpha$, solve the system of equations and interpret the results. In your case you can even neglect the linear motion portion of the problem and concentrate entirely on the rotational part. $\endgroup$
    – dmckee --- ex-moderator kitten
    Apr 6, 2015 at 2:27

2 Answers 2

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After further research, I found that the moment of inertia of a system consisting of multiple objects, like the arm-bowling ball system in the problem, can be found simply adding the moments of inertia of each object 1, though use of the horizontal axis theorem may be necessary if the object is rotating around an axis parallel to its typical axis of rotation 2.

The moment of inertia of the arm is given, and the moment of inertia of the bowling ball can be modeled as the moment of inertia of a point mass, given by the equation: $I=MR^2$. Since the moment of inertia of a composite object is the sum of the moments of inertia of its parts, the rotational inertia of the arm-bowling ball system is:

$I_{system}=0.48$ $kg$ $m^2$$ + $($2.7$ $kg$)⋅($0.64$ $m$)$^2$$=1.6$ $kg$ $m^2$

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Calculate the MI (moment of inertia) of the bowling ball around the axis of rotation , i.e - shoulder.
Assuming the bowling ball to be a solid sphere (it is actually a shell?), since the shell thickness isn't given, this would be-
Icm = (2/5)MR^2 (MI about it's center of mass)
Hence MI around shoulder, using parallel axis theorem -
Ishoulder = Icm + Md^2 ( where d is the displacement between the center of mass of the bowling ball and the shoulder-axis of rotation)
The moment of inertia of the arm is given and this can be directly added to Ishoulder.
Itotal = Ishoulder + Iarm
Now you have MI total. All the torques are known as well (arm weight and bowling ball weight and their relative displacements from the shoulder).
You should be able to take it from here

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  • $\begingroup$ On second reading, the radius of the bowling ball isn't given and hence should be assumed to be a point mass. I total = .48 + 2.7* (6.4)^2 $\endgroup$
    – chilljeet
    Apr 7, 2015 at 5:05

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