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$\bullet$ Use the formulas of Table 9.2 to find the moment of iner-tia about each of the following axes for a rod that is 0.300 $\mathrm{cm}$in diameter and 1.50 $\mathrm{m}$ long, with a mass of 0.0420 $\mathrm{kg}$ : (a) about an axis perpendicular to the rod and passing throughits center; (b) about an axis perpendicular to the rod and pass-ing through one end; and (c) about an axis along the length ofthe rod.
a) $7.88 \times 10^{-3} \mathrm{kg} \mathrm{m}^{2}$b) $3.15 \times 10^{-2} \mathrm{kg} \mathrm{m}^{2}$c) $4.7 \times 10^{-8} \mathrm{kg} \mathrm{m}^{2}$
Physics 101 Mechanics
Chapter 9
Rotational Motion
Physics Basics
Rotation of Rigid Bodies
Dynamics of Rotational Motion
Equilibrium and Elasticity
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throughout this problem, I'll be referring to Table 9.2 that the problem is also referring to. And so for part A. We need the moment of inertia represented by this here one twelve 12 and we'll squared. That is, for the baton spending across an access like this. It's moving like this, so pointing in these values that they give us, we get that eye is equal to hold on. Let me racist we get that eye is equal to 7.88 times 10 to the negative three. This is in units of kilograms, meter square for probie. Our access is actually here on the end, and it's rotating this way. And so in this case, the table tells us that the moment of inertia is 1/3 and I'll square like that. And so this implies that eyes eagle two 3.15 times tend to donate to killed ransom. You're squared just pulling in the values that they use in the problem from Marci, we're going down the length of the cylinder. So if this is looking down the length of the cylinder and this is the access right, this is like this and the access is running through it. Now the thing is rotating this way, just like this. And so the table now tells us that I is equal to 1/2 Imar squared and then playing the values this time gives us eyes equal to 4.72 times 10 to the negative eight kilograms meter squared like that.
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