Alright, thanks for the suggestion, I'll try to stick to that from now on.Here is the numerical correction, I = 1.28 + 0.64 = 1.92.
Now uhm.
The center of mass of the disk is the center of the disk, which is rotating alone point A.
Radius would be OP/2. It'd move pi/2 radians, shoudln't it...
Alright, so the center of mass would be the center of the disk itself.
So it'd be..
1/2 MR^2 + MD^2.
Where the distance here would be OP/2.
So 1/2 (4kg) * 0.8m + 4kg*0.4m = 3,2.
Now I found the moment of inertia (if I did it right, lol) of the disk rotating alone point A.
The moment that...
I am sorry if I am a little slow but as I said I am just starting out with these.
The disk isn't rotation around its center of mass (the center), so we have to apply the integral of r^2dm
in order to find out the moment of inertia?
Or maybe I could apply the parallel axis theorem?
Holy, that's kind of unexpected, I mean I don't see any wire nor talking about gravity, so yeah, the idea didn't even pass by my mind.
I don't get why you say that "This is all about rotation about a point other than the center"
Why's that?
Homework Statement
A system is made of the following things:
A homogeneous disk of mass m = 4kg.
Having radius r = 80cm's.
A point P (m = 2kg) free to rotate around a horizontal axis which is perpendicular to the disk in point A.
AO = OP/2
If the system starts from rest and AP forms a...
the spring would be perpendicular to the inclined plane, something like this (check attachment).
So, a question, when we have an inclined plane, and we want to find the potential energy.
mgh
h is always the height from the ground or it should be relative to the inclination of the plane?
It means that when it is not compressed, the spring has a length of 30cm's.
Then when the block pushes it, it's compressed of a maximum of 12 cm's.
Now I would like to know what I must plug into x.
Homework Statement
A body of mass m = 2.5kg is on a frictionless inclined plane (by 40°). At an height of h = 1.6m.
On the bottom of the plane there's a spring, which at rest has an extension of 30cm.
If the spring is compressed maximum of 12cm, what's the "k" of the spring?
Homework...
Ok so:
1/3ML^2 => 1/3 0,5kg * (1.50)
I = 0,25.
mgh = 1/2Iw^2
w = square root of... (2mgh)/I
So
w = 5,42 rad/sec
Velocity = radius * w => velocity = 1,50 * 5.42 = 8.13m/s
And then the other stuff is just an elastic collision which I should know how to handle.
Just one...
It'd be 1/2Iw^2
where
I = momentum of inertia
So it'd be:
1/3ML^2 => 1/3 0,5kg * (0.75)
(I am not sure, but this should be it?)
Then I can do the following:
mgh = 1/2Iw^2
Where I have I, m and g.
I can assume that h = 0,75 because it'd be the radius, and so I could find the angular velocity?
Since there is no talking about density I can assume the center of mass is just L/2 so:
x = 0,75.
Right? Now I understand that I have to do something with energy, so like:
mgh = 1/2mv^2
But I have to decompose the motion in x and y. I am not sure if this is correct but this is my...
Homework Statement
"A bar line AB has a length of 1,5m and a mass of 0,5kg.
It is stuck in the point B in a horizontal fixed hub, and it can oscillate freely with no friction in the verical plane.
at time t=0, the bar line, which was still (not moving, inertia), is set free to rotate.
When it...