You use the exact same lecture slides that my professor uses. I dont know if that's legal for him? anyways, i could watch all your videos instead of going to lecture. Its the exact same thing!!!! and you explain everything better. Thank god i found this channel
Dear Professor Selleck, thank you for you video. I just don t understand why didn't you apply additionally to inertial times alpha the moments induced by (a)t of the rod and (a)t of the sphere as the Sum (Mk) on the right hand side of the equation?
Sir, if a rigid body rotates about a fixed axis then, are all the points on the axis of rotation have no acceleration ? And if a point on the axis of rotation doesnot have acceleration , i think there is an error in the book when deriving the equation of moment about the axis of rotation in section 17.2 page 410 ..as you will see the book will treat point p as if it had an acceleration .
Section 17.2 is about rigid bodies rotating and accelerating. There is no fixed axis of rotation in the derivation. Fixed axis rotation is in section 17.4.
for the second example , sum of moment ................... me and my friends and the whole people on the earth believe that the IG for rod = ml^2 divided by 12 with all respect
osama mehdawi It is true that the moment of inertia of a homogeneous rod about its center is 1/12 * m * l^2, but I am summing moments about the rod's end (point O), and that moment of inertia is 1/3 * m * l^2.
+osama mehdawi If you were to use the equation mI^2 for IG, you would have also had to apply the parallel axis thereom of mr^2, but to avoid that, he took the moment of inertia at the end. If you were to do it using the other method, you would get the same exact result. Also for someone that claims that he is addressing someone with respect, you seem to show little proof.
And this is way when you go to school and don’t pay attention, you might generally go around with misconceptions. Besides on the video he clearly said about its end. You could use your formula and the paralleled axis theorem to get the same answer. Inertia is about a rotating axis in this case O
@@brandonphan1437 I assume you are talking about example 1 at 7:12. The mass MoI of a rod about G ( the rod center) is 1/12 * m * l^2, not 1/3 * m * l^2.
You use the exact same lecture slides that my professor uses. I dont know if that's legal for him? anyways, i could watch all your videos instead of going to lecture. Its the exact same thing!!!! and you explain everything better. Thank god i found this channel
I wish I found this channel sooner!
Thanks for the kind words!
Dear Professor Selleck, thank you for you video. I just don t understand why didn't you apply additionally to inertial times alpha the moments induced by (a)t of the rod and (a)t of the sphere as the Sum (Mk) on the right hand side of the equation?
I believe I did that.
Sir, if a rigid body rotates about a fixed axis then, are all the points on the axis of rotation have no acceleration ? And if a point on the axis of rotation doesnot have acceleration , i think there is an error in the book when deriving the equation of moment about the axis of rotation in section 17.2 page 410 ..as you will see the book will treat point p as if it had an acceleration .
Section 17.2 is about rigid bodies rotating and accelerating. There is no fixed axis of rotation in the derivation. Fixed axis rotation is in section 17.4.
Ok .thank you sir
14:00 when you back sustitute to get Ot how come you didn't take into consideration the negative sign for alpha.
You are correct. Ot is -399 N.
for the second example , sum of moment ................... me and my friends and the whole people on the earth believe that the IG for rod = ml^2 divided by 12
with all respect
osama mehdawi It is true that the moment of inertia of a homogeneous rod about its center is 1/12 * m * l^2, but I am summing moments about the rod's end (point O), and that moment of inertia is 1/3 * m * l^2.
+osama mehdawi
If you were to use the equation mI^2 for IG, you would have also had to apply the parallel axis thereom of mr^2, but to avoid that, he took the moment of inertia at the end. If you were to do it using the other method, you would get the same exact result.
Also for someone that claims that he is addressing someone with respect, you seem to show little proof.
And this is way when you go to school and don’t pay attention, you might generally go around with misconceptions. Besides on the video he clearly said about its end. You could use your formula and the paralleled axis theorem to get the same answer. Inertia is about a rotating axis in this case O
@@ColinSelleck Using 1/3 * m * l/2 with values of mass = 15kg and l being 0.9l you end up with 4.05. Rather than 1.35
@@brandonphan1437 I assume you are talking about example 1 at 7:12. The mass MoI of a rod about G ( the rod center) is 1/12 * m * l^2, not 1/3 * m * l^2.
Good video but you sound SO unenthusiastic. I could fall asleep listening to this. I wish there wasore energy and excitement. Good lecture nonetheless