A dumbbell consists of two equal compact masses of 2 kg each, connected by a massless rod of length 0.8 metres.

What is the moment of inertia of the dumbbell about an axis through the centre of, but perpendicular to, the rod?

could someone just tell me how to get started on this as i'm completely lost and ive an exam in three weeks:(

4 answers

add up the m r^2
the center of mass is in the middle.
Therefore r for each = 0.4 meters
For each of them m r^2 = 2*.4*.4 = 0.32
So the total I = 0.64
thats great thanks very much

(ii) Using the parallel-axis theorem or otherwise, find the moment of inertia of the dumbbell about an axis through one of the masses but perpendicular to the rod

and for this question i need :
I = Ic.m. + Md2
thats right isnt it?
Yes, M is the total mass and d is .4 meters
(iv) A tangential force of 20N is applied to one of the masses in the plane of rotation of (iii) above for 3.0 s . (Both the instantaneous velocity of the mass and the applied force are pointing in the same direction at all times). What is the final angular velocity of the rod?

for this will i need to use te formula for torque?
i can get very little information from this question any help you could give qould be great