For anyone confused about the last one why it doesn't have translational KE, it's because the axis of rotation of the rod is not moving. Since that isn't moving it wouldn't have the 1/2mv^2 term.
I don't understand your question. But it's initially at rest so it only has potential energy from a height of .5m. When it reaches the bottom of it's rotation all of it's initial potential energy is converted into rotational energy. If it's not spinning initially then it doesn't have rotational energy.
Could u do a problem with masses initially at the same height?Then find the angular velocity of the pulley after a heavier mass a fallen through some distance,h.
+Vadi Eghterafi Yes it does. The center of mass will have kinetic energy r*w. Where r is 1/2 meter. Edit: Actually he used the parallel axis theorem, it is the same as using the I about its center and then adding the linear momentum as a separate term. Try it out and check for yourself. 1/12 + 1/4 = 1/3.
In the E', since the potential energy of the smaller mass would be acting in the opposite direction of the velocity, shouldn't you subtract the M2gh instead of add it?
That isnt a -h, it is just h. To be negative, it would have to go below the ground. Use your logic, not what you physicians make you memorise, like negative and stuff.
In the first example, if there is no sliding between the rope and the cylinder that means there is a frictional force component? shouldn't that be addressed in the energy equation? and in the final example the inital energy of the system (i.e. P.E.) is mg(1) not mg(0.5), because the ruler is on its horizontal position and gravity will act on it fully(no need for centre of mass-can just assume its a mass point) for a full 1 metre before it will hit/touch the ground. After the rotation when K.E. is highest and P.E. = 0, the moment of inertia used is that of a circle, where as the example uses a ruler (rectangular shape)and rotation of axis is edge and not centre so assuming its a thin ruler then I= (m/12) * (h^2+w^2). Sorry my adhd got me ranting and ranting xD but it's just safer to use the example of a yo-yo ball/string or point mass lol, Nice video btw keep it up!
+Ahmed Islam You are misunderstanding what the frictional force is meant to be. It is not "work" in the sense that the work done from point 1 to point 2 is Force*Distance, but instead compare it to simpler dynamics problems. In simple work-energy problems the pulley is considered mass-less and friction-less, but in this more advanced problem the friction force is causing the pulley to rotate, instead of just sit there and allow the rope to slide. The force of the friction is what's causing the rotation to occur, it is already being accounted for. As to your second point. His reference point is halfway up the meter stick when in the vertical position. So that PE)g in position 2 is zero. Remember, the CHANGE in height is whats important, reference points are arbitrary. As for your final point, .5*I*w^2 is the equation for all rotational motion, whether it's a bar rotating about a point or a circle rotating about its center. His I equation of (1/3)*m*L^2 is simply the I of a bar rotation about its end point. He would achieve the same end result if he instead used the I for a bar about its central axis, (1/12)*m*L^2 and then added the linear kinetic energy term in after, (1/2)*m*v^2, where v=r*w and r=L/2. Hope this helps some of your confusion!
The correct answer for me had m_1 on the left hand side of the equation equal (m_2-m_1) and it didn't have m_2gh on the right hand side of the equation. Here's the page where I found it felix.physics.sunysb.edu/~allen/131-06/HW/EoC7.pdf
There is no kinetic energy in a moving mass there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers
if we consider the rope to not stretch, then due to its uniformity, it would be impossible for any section of the rope to move faster than the one before/after it. So the speed (in an ideal case where the rope does not bend or slip) will be the same at every point.
Can u explain why u are using conservation of energy? Because I thought conservation of energy is used only when there is no non-potential force or when nonpotential force is perpedicular to direction of object
For anyone confused about the last one why it doesn't have translational KE, it's because the axis of rotation of the rod is not moving. Since that isn't moving it wouldn't have the 1/2mv^2 term.
I LIKE THIS VIDEO, IT HAS BEEN REALLY HELPFUL
It's mr^2 for a point mass that is a distance r from an axis. But it's 1/2 mr^2 for a solid cylinder that is rotating about its center of mass.
I don't understand your question. But it's initially at rest so it only has potential energy from a height of .5m. When it reaches the bottom of it's rotation all of it's initial potential energy is converted into rotational energy. If it's not spinning initially then it doesn't have rotational energy.
In the last example, why can we not say the rod has translational KE as well ?
The axis of rotation is not moving for the last rod.
Could u do a problem with masses initially at the same height?Then find the angular velocity of the pulley after a heavier mass a fallen through some distance,h.
In your final example, doesn't the meter stick have some regular, linear kinetic energy because the center of mass is changing in its position?
+Vadi Eghterafi Yes it does. The center of mass will have kinetic energy r*w. Where r is 1/2 meter.
Edit: Actually he used the parallel axis theorem, it is the same as using the I about its center and then adding the linear momentum as a separate term. Try it out and check for yourself. 1/12 + 1/4 = 1/3.
In the E', since the potential energy of the smaller mass would be acting in the opposite direction of the velocity, shouldn't you subtract the M2gh instead of add it?
That isnt a -h, it is just h. To be negative, it would have to go below the ground. Use your logic, not what you physicians make you memorise, like negative and stuff.
I watched almost all people who give physics tutorials. your questions are the best but the sound that your pen generating is disgusting
lol the second problem you did was the exact one that came on my test.
If only my school teachers could slow down to a comprehensible pace like yours
In the first example, if there is no sliding between the rope and the cylinder that means there is a frictional force component? shouldn't that be addressed in the energy equation?
and in the final example the inital energy of the system (i.e. P.E.) is mg(1) not mg(0.5), because the ruler is on its horizontal position and gravity will act on it fully(no need for centre of mass-can just assume its a mass point) for a full 1 metre before it will hit/touch the ground. After the rotation when K.E. is highest and P.E. = 0, the moment of inertia used is that of a circle, where as the example uses a ruler (rectangular shape)and rotation of axis is edge and not centre so assuming its a thin ruler then I= (m/12) * (h^2+w^2).
Sorry my adhd got me ranting and ranting xD but it's just safer to use the example of a yo-yo ball/string or point mass lol,
Nice video btw keep it up!
+Ahmed Islam You are misunderstanding what the frictional force is meant to be. It is not "work" in the sense that the work done from point 1 to point 2 is Force*Distance, but instead compare it to simpler dynamics problems. In simple work-energy problems the pulley is considered mass-less and friction-less, but in this more advanced problem the friction force is causing the pulley to rotate, instead of just sit there and allow the rope to slide. The force of the friction is what's causing the rotation to occur, it is already being accounted for.
As to your second point. His reference point is halfway up the meter stick when in the vertical position. So that PE)g in position 2 is zero. Remember, the CHANGE in height is whats important, reference points are arbitrary.
As for your final point, .5*I*w^2 is the equation for all rotational motion, whether it's a bar rotating about a point or a circle rotating about its center. His I equation of (1/3)*m*L^2 is simply the I of a bar rotation about its end point. He would achieve the same end result if he instead used the I for a bar about its central axis, (1/12)*m*L^2 and then added the linear kinetic energy term in after, (1/2)*m*v^2, where v=r*w and r=L/2.
Hope this helps some of your confusion!
you are such a good teacher !!!!!!!!!! LADY BUGS
🐞
Ur amazing man thank u God bless u
Why is it not mg = .5Iw^2 + mg(.5) ??
isnt I = mr^2?
not 1/2 mr^2
nop, what he did was right and its already been 8 years.....hope you're not working at McDonald's because of that
The correct answer for me had m_1 on the left hand side of the equation equal (m_2-m_1) and it didn't have m_2gh on the right hand side of the equation. Here's the page where I found it felix.physics.sunysb.edu/~allen/131-06/HW/EoC7.pdf
These videos are great! :D
Bless you
There is no kinetic energy in a moving mass there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers
why they all have the same velocity ?how can we prove that mathematically!
if we consider the rope to not stretch, then due to its uniformity, it would be impossible for any section of the rope to move faster than the one before/after it. So the speed (in an ideal case where the rope does not bend or slip) will be the same at every point.
Can u explain why u are using conservation of energy? Because I thought conservation of energy is used only when there is no non-potential force or when nonpotential force is perpedicular to direction of object
because there is no other force acting on the object so the system is isolated. that is the cause of conservation of energy.
Would you please use a pen instead, the sound of the maker on the page is quite irritating.
whatttttt lmaoooo
stfu and be grateful
omg thank you soo much @_@ I was lost before this xD
v nice
please use a different marker in your next videos.
No.
that pen sound so disturbing