classical physics - rod anchored to ground, rotational inertia

[lifes3ps]

going over classical physics for my physics GRE and im forgetting the easy stuff.

problem 1: Rod length L mass M has 1 end anchored to ground, the top is allowed to freefall, whats the speed of the tip when hits ground?

problem 2: 7 pennies (uniform, equal disks) are arranged in a hexagonal planar pattern. whats the moment of inertia around the axis perpendicular to the plane of the pennies, thru the center penny

i have the answers as its from a gre test book but any help helpful

thanks

[Schweitzer]

I'm taking the Physics Subject test on Nov. 3rd as well.

For the first problem I just use conservation of energy. Initial Potential energy = Final Kinetic Energy
M*g*(height of center of mass) = .5 * M * V^2
Height of center of mass is obviously L/2
V=[(g*L/2)*2]^.5

For the second I just assume pennies are point masses and sum:
I=mr^2
Since all m's and r's are equal (in magnitude) then
I=6mr^2

[Schweitzer]

On further thought I was too quick on the second; you can't assume point masses.
Moment of Inertia of a disc is .5*m*R^2 (rotating perpendicular to plane disc is on)

I can't remember how to add moments of inertia so I guess I would just multiply them all by their distance from the center squared and sum. Not sure if that is right though...

I would get .5MR^2 + 3MR^2*r^2
R is radius of penny, r is distance of penny from center

[lifes3ps]

schweitzer, im taking the the 3rd too, gl.

both are wrong. im beginning to think it could be rotational...for the first...you cant take the hight of the center of mass cuz it just wants the tip of the rod and theyll be moving at different angular velocities. if the w=v/r. im thinking if you can scale that out then maybe.


the answers (3gl)^.5

and #2 was (55/2)* mr^2

[lifes3ps]

in your #2 r=2R btw

[Schweitzer]

[ QUOTE ]
schweitzer, im taking the the 3rd too, gl.

both are wrong. im beginning to think it could be rotational...for the first...you cant take the hight of the center of mass cuz it just wants the tip of the rod and theyll be moving at different angular velocities. if the w=v/r. im thinking if you can scale that out then maybe.


the answers (3gl)^.5

and #2 was (55/2)* mr^2

[/ QUOTE ]

Yeah you're right I was way off.
My potential was fine but kinetic energy should change from .5*m*v^2 to its rotational equivalent: .5*I*w^2
then:
MgL/2 = .5*I*w^2 since w=v/r then
MgL/2 = .5*I*(v/L)^2 -> V = [M*g*L^3/I]^.5
I had to look up I of rod and found I=1/3*m*L^2
then you get V=[3*g*L]^.5

Not sure about the second question but I'll try and figure it out.

[TomCowley]

I'm sure there's an easier way for the first one, but using the initial PE = final KE method, initial PE is 1/2 Mg. Since velocity at the end is proportional to distance from the anchored part of the rod, we can call velocity at distance r kr
Final KE = Int [r on 0,1] 1/2 M (kr)^2 dr = 1/6 M k^2 r^3 evaluated at 1 and 0 = 1/6 M k^2. So 1/6M k^2 = 1/2 M g.

So 1/3 k^2 = g, k=(3g)^0.5, so velocity at the end = k*1 = (3g)^0.5 which is the same answer lifes3ps got.

[Siegmund]

For the moment of inertia one... the moment of inertia of an off-axis penny is the penny's moment of inertia around its own center, plus the moment of inertia of a point mass at that distance from the central axis.

So, final answer will be 7*MI of penny around its own axis + 6*MI of point mass one penny-diameter from the axis of rotation.

[Schweitzer]

[ QUOTE ]
problem 2: 7 pennies (uniform, equal disks) are arranged in a hexagonal planar pattern. whats the moment of inertia around the axis perpendicular to the plane of the pennies, thru the center penny


[/ QUOTE ]

Ok, got this one. Moment of inertia of a disk =.5*M*R^2
Parallel Axis Theorem: I = Icm + m*r^2
Icm is moment of inertia about center of mass.

Then, I = .5*M*R^2 + 6*(.5*M*R^2 +M*r^2)
I = 7/2*M*R^2 + 6*M*r^2
plugging in r=2R
I = 7/2*M*R^2 + 24*M*R^2
I = 55/2*M*R^2

[lifes3ps]

sweet, thanks schweitzer....pm me or something with your aim info...so we can bounce [censored] off each other, that was the last of the classical problems i went thru yesterday, im doing thermodynamics tonight.

off to library