The rigid body shown is rotated about an axis perpendicular to the paper and thr
ID: 2243547 • Letter: T
Question
The rigid body shown is rotated about an axis perpendicular to the paper and through the point P. If M = 0.40 kg, L = 60 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 48.0 rev/min? Neglect the mass of the of the connecting rods and treat the masses as particles.
The rigid body shown is rotated about an axis perpendicular to the paper and through the point P. If M = 0.40 kg, L = 60 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 48.0 rev/min? Neglect the mass of the of the connecting rods and treat the masses as particles.Explanation / Answer
KE=.5(Inertia)(omega^2)
Since we dont know I then we have to find.
I= .4kg*.6m^2 + .4*.3m^2 = .144 + .036 = .18 (we dont include the M in the center because it doesnt have length therefore there is no moment of inertia on the object.
===> KE= .5(Inertia)(omega^2)=> .5(.18)((48*2*pi/60)^2)= 2.3 Joules
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