Recent content by KataruZ98

Suppose for the sake of the argument that the two are connected, not simply placed one below the other.

The total kinetic energy of a double pendulum can be calculated through the formule reported in the following article: https://dassencio.org/33 This works if the double pendulum in question is formed by two masses connected to each other and — one of them — to the point of origin by a...

Well, to be openly clear, I’m using this figure as a model to find the MOI of a forearm about the shoulder, when there’s a rotation happening both at the elbow and shoulder. I’d like to see the resulting energy and torque calculated, but I need the MOI of the forearm around the shoulder to...

As I said in my first reply, the rotation I’m talking of can be thought as that of the Moon around the Sun — with the former already going circles. Here you can think at the cylinder as the Moon, the first axis as Earth and the second axis as the Sun. I hope I made my case clear.

To give a visual example from RL to clarify what I meant, take the Moon’s orbit around the Sun and replace the satellite with the specified cylinder (and of course scale down the distances and sizes as described in the question). Pretty much you could replace my question with “how can the Moon’s...

Considering a cylindrical rigid body of length 3 m and wide one. The body is rotating about an axis passing through one of its bases and perpendicular in respect to the length. At the same time, the same cylinder is orbiting about another axis parallel to the first — but distanced 10 m from the...

If you don’t mind switching up, could we focus on the case of the cylinder hitting like a pendulum the sphere - as it was my original query? Anyway though I must thank you for the explanation, and I apologize for having sidetracked. EDIT: we can also treat it as a 10m long, 2m wide cylinder if...

Where did that 1.67 come from? It is not equal to the moment of inertia of the cylinder. I mistakenly used the formula for the moment of inertia of a cylinder about an axis perpendicular and passing through the center of mass, then used the parallel axis theorem to get the moment of inertia by...

Alright, that’d be: I(c)*w(c)+I(s)*w(s) Or: 1.67*1.57+0.17*0

I see. Say, Vc is 1m/s, would that suffice?

Alright, the total linear momentum would be: Mass(cylinder)*Initial Velocity(cylinder)+Mass(sphere)*Initial Velocity(sphere) I didn’t give an initial linear velocity for the cylinder I realize, so if we call that X the formula would be: 5X+0

As far as I know, momentum is always conserved - be it angular or linear. However mechanical energy isn’t in an inelastic collision.

I see, that’d be momentum and energy if IIRC

I see. Apologies, but would you guide me through my errors to see what went wrong in my assumptions? EDIT: just saw what you’ve add. Could you explain more in depth then the mechanism of KE transfer in this situation?

Well isn’t torque of the rotating body equal to its KE divided by the angular displacement (90 degrees for the question’s purpose)?