Calculate the moment of inertia (in kg·m2) of a skater given the following information.
(a)
The 64.0 kg skater is approximated as a cylinder that has a 0.120 m radius.
_____kg·m2
(b)
The skater with arms extended is approximately a cylinder that is 56.0 kg, has a 0.120 m radius, and has two 0.950 m long arms which are 4.00 kg each and extend straight out from the cylinder like rods rotated about their ends.
oubaas
a)
MoI J = m/2*r^2 = 64/2*12^2*10^-4 = 0.46 kg*m^2
b)
MoI J' of each arm :
Steiner theorem shall apply :
J' = m*L^2/3+m*r^2 = m(L^2/3+r^2) = 4*(0.95^2/3+0.12^2) = 1.26 kg*m^2
computation of total MoI J'' :
J''= J+2*J' = 2.52+0.46 = 3.0 kg*m^2
Steve4Physics
For (a) use the standard formula I = ½MR² = ½*64.0*0.120²
For (b) the wording is ambiguous because it is not 100% clear if the inner end of each arm is taken to be on the axis of rotation or on the surface of the cylinder. The wording implies the former, giving:
Each arm is treated as rod rotating about its end: I = ⅓mL². You add the moments of inertia of the cylinder and the two arms:
I = ½MR² + 2(⅓mL²) = ½*56.0*0.120² + 2(⅓*4.00*0.950²)
You can complete the arithmetic. Remember to round answers to 3 sig. figs.