derive I=mR^2(gRF/aTr-1) derive I=mR^2(gRF/aTr-1) Finally we can use the relatio

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derive I=mR^2(gRF/aTr-1)

derive I=mR^2(gRF/aTr-1) Finally we can use the relationship between torque and angular acceleration, Eq. (3), and Eq. (7), to solve for the moment of intertia in terms of quantities we can measure: (8) Prelab Questions I. Derive Eq. (8). 2. Calculate the moment of inertia for two point masses M equidistant from the center of rotation. (This is the simplest model of the dumbbell system). Will the true moment of inertia of the dumbbell system be larger or smaller than this estimate?

Explanation / Answer

First you start by defining the axis of rotation. Is it vertical along the center line exactly half-way between the center of mass for the two cylinders? Or it is around the longitudinal axis running through the middle of the two cylinders and the handle in between them? I = kMR^2 will be different between the two cases because k will be different and R will be different. M, the mass of the two cylinders, will be the same either way.

For the first case, with the mass of each cylinder as m, I'd use the two point mass model I = m*m/(m + m) * L^2 = m^2/2m * L^2 = 1/2 mL^2 where L is the center to center length of the handle between m and m.

For the second case, along the longitudinal axis, I'd ignore the handle mass and go with I = 1/2 mR^2 + 1/2 mR^2 where m is the mass of each cylinder and R is the radius of each one.

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