PhanthomJay said:Equation 1 of the solution manual is correct, but I am not in favor generally of using inertial pseudo forces. I also don't understand your attempt. Personally, I find it simpler to take a free body diagram diagram of the mass and apply Newton's 2nd Law for translation to solve for the tension in the portion of the cord attached to the weight . Then use the net torque equation taking a FBD of the pulley.
Yes, that seems like the easier way.mech-eng said:thanks for the answer.my teacher used your method as well.
that is correct, but i am not sure why I said inertial forces were used since I can't open up your attachment anymore. It may have been an incorrect conclusion on my part.But what is inertial pseudo forces.
I think inertial forces are imaginary and we add them to the system while implying D'lambert
principle for taking the system to equilibrium.
yes, the tension is not equal to the block weight when acceleration is involvedmeanwhile my attempt was wrong and ı thought
tension in that portion is equal to the weight of the block.
well I guess you could use that approach where the inertia of the hanging mass, treated as a particle, is mr^2, where r is the pulley radius and m is the mass of the block; then you look at the entire system without including the internal tension force, where the sum of torques , mgr -F_s(r), is equal to the Iα of the pulley plus the Iα of the hanging mass, where the Iα of the hanging mass is mr^2(α), which maybe is what the book solution was doing, to get the same result, but this is not a good way of doing it, in my opinion.my another question about this equation
is can we use moment of inertia of the block in spite of its not turning like we a mass turning
free end of a pendulum.have a nice day.
mech-eng said:well I guess you could use that approach where the inertia of the hanging mass, treated as a particle, is mr^2, where r is the pulley radius and m is the mass of the block; then you look at the entire system without including the internal tension force, where the sum of torques , mgr -F_s(r), is equal to the Iα of the pulley plus the Iα of the hanging mass, where the Iα of the hanging mass is mr^2(α), which maybe is what the book solution was doing, to get the same result, but this is not a good way of doing it, in my opinion.
But block is doing a translational motion and moment of inertia is only for rotational motions but
there is a relation between them in that problem.Can you discuss with other mentor friends.
I see you are not convinced. It is not an easy concept to grasp. Try to avoid using it when practicable. It stems from the definition of moment of inertia, and the calculus behind it.mech-eng said:But the example you gave is that of one degree of freedom i.e there is only one independent
motion.because rotatinal motion of the pulley and translational motion of the block are dependent to each other (can we say they are actually the same? ) your approach is true but it shouldn't be always true.
Fair enough, that is good insight. You should find that an object with linear momentum also possesses angular momentum about a chosen axis, in the same way that an object with linear acceleration also possesses angular acceleration about a chosen axis. Even when there is not rotation per se. Perhaps it may be helpful to replace the word "linear" with "tangential".mech-eng said:But an always falling object can't make an angular motion about an axis.I am not sure but this concept might related with linear and angular momentum.Let's postpone this after I worked momentum.