- #1
Alex126
- 84
- 5
Homework Statement
There is a block of known mass m on a horizontal surface. The block is connected to a pulley via a rope. The other end of the rope is pulled vertically, downwards, with a known force F. The pulley has a known moment of inertia I and radius r.
Calculate the acceleration a of the block.
Homework Equations
ΣM = I*α
α = angular acceleration = a/r (a = tangential acceleration)
ΣF = m*a (F = forces acting on a body)
M = r*F
The Attempt at a Solution
This would be the drawing, though I'm a bit unsure about that Tension force T:
Anyhow, I wrote the equation for the forces acting on the body, ignoring the Y axis since it's balanced:
+T = m*a
Then there is the other part of the problem, which focuses on the angular momenti on the pulley. Since the rotation is obviously clockwise, I'll call + the clockwise one, and - the counterclockwise one. Now, I know that there is a MF, since clearly that force exists and so does its momentum, but is there also another force to consider, namely the tension T? If I think about it, there shouldn't be any other momentum, since that force F is the only actual force applied to the pulley, BUT surely the same force would make the pulley rotate differently if the mass m was also different: no mass -> same F should make the pulley go "faster"; plenty enough mass -> no motion at all.
Instinctively, I thought there should be something like this:
+MF - MT = I*α
Rationally, I don't know where this "-T" comes from. The "T" I drew quite clearly goes in the same direction as F, so if anything it should be +MT. But then again, on the bottom part of the rope (where I drew F) maybe there is also a counter-Tension T2 (equal in magnitude to the T I wrote, but pointing upwards, i.e. opposing the motion), and that's the one producing that -MT.
The procedure after this point is straightforward, since I can just use the various equations with MF = r*F and MT = r*T, and a = α*r, combine them all in a system with T and α and get to the solution. What goes in the system, that's the question.