Moment of Inertia of a grouping of pennies

[EventHorizon4]

Hi,

An old GRE problem asks what the moment of inertia of seven pennies, arranged in a hexagon with one in the center, all touching each others' edges is, about the axis that passes through the center of the central penny and is normal to the plane of the pennies.

Each penny is a uniform disc of mass m and radius r, and the answer is apparently 55mr/2.

Can anyone help with this?

Thanks

 

[Doc Al]

Where are you stuck?

 

[EventHorizon4]

I obviously know how to calculate the moment of inertia of a disc, but I just don't understand how to account for multiple discs all centered at different places in a plane. It doesn't seem like it can be one integration...that would be really ugly. So is there some sort of way to "add" the moments of each disc together? That's what I don't understand.

Sorry I haven't gotten very far...

 

[Doc Al]

Hint: All that matters is how far each disk is from the axis. No integration needed.

 

[pgardn]

I obviously know how to calculate the moment of inertia of a disc, but I just don't understand how to account for multiple discs all centered at different places in a plane. It doesn't seem like it can be one integration...that would be really ugly. So is there some sort of way to "add" the moments of each disc together? That's what I don't understand.

Sorry I haven't gotten very far...

Do you know how to find the moment of inertia of two pennies whos centers are a distance r away from each other and the axis of rotation is right between them? hint assume the center of mass of each penny is a point in the middle of the penny...

oopss I see I interposted, nevermind.

 

[EventHorizon4]

Wasn't familiar until now with the "parallel axis theorem," but that seems to have been the ticket here. Using that to calculate the moment of inertia for the pennies NOT rotating on their center of mass, then adding all seven together, gets the answer. Thanks.

 

[Doc Al]

Yep, the parallel axis theorem is what you need.