At 14:40, could you please explain why the acceleration a(G) is towards the right? Shouldn't it be towards the left since the angular acceleration is CCW?
@@younamsayin No. The alpha is wrt to truck and it can be CCW but the mass center can be accelerating to the right. Look at the relative acceleration equation.
@@ColinSelleck i'm pretty confused here... I thought the motion of this truck will cause the ring to move towards the left. IF we were to use angular velocity instead of angular acceleration, and velocity instead of acceleration, then in this scenario, would the angular velocity become CCW and the velocity to the left?
@@younamsayin The pipe does move to the left wrt to the truck. But the absolute acceleration of the ring is to the right because the truck is accelerating.
Thank you so much for making and uploading these videos, they have been extremely helpful. I am having difficulty understanding why in he firs example, you used the inner radius instead of that of he body.
I use the inner radius because the wheel is slipping at the contact point with the ground. You can visualize the cable being wrapped around the inner radius as the wheel moves to the right and the instantaneous center of zero velocity is at point A..
In 8:32 the spool problem, does anyone know why it moves to the right? I think it should move to the left because the moment is counterclockwise. Thanks
At 7:57 you tell us about the acceleration of the mass centre being consistent with the angular acceleration. Can you please elaborate on this, I'm not sure exactly sure about what this means. Thanks
This statement is only true for frictional rolling problems that I had just gone over in the video. If there is no slip (and a homogeneous cylinder so the mass center is at the center of the cylinder), then Ag = r times alpha. If you assume alpha is CCW, then Ag must be assumed to the left and vice versa.
Colin Selleck Thank you so much. If it's not a rolling problem, then can I relaxed with assumed directions of alpha and acceleration of the mass centre?
First thank you for the video. If the spool in your first example is 1962 [N] / 9.81 = 200 [Kg], Not 20 [Kg] there for your solution is incorrect. Unless I missed something.
Hello guys. I attempted fundamental problem 17.13 but saw that the mass of rod wasn't accounted for in the sum of forces in the y-direction, is there some fundamental concept I'm missing or is it just an error on their side(highly unlikely)?
This confused me too at first until I realized the drawing is a plan view (from the top), not a side view. So gravity does not play a role. And since the plane is smooth, neither does friction.
AT 5:37 I didn't say not to use F=(mu)N. I said first assume no slip (so F is not equal to (mu)N) and solve for the frictional force. Then compare the friction with (mu)N. If it is greater, then use F = (mu)N.
At 14:40, could you please explain why the acceleration a(G) is towards the right? Shouldn't it be towards the left since the angular acceleration is CCW?
The pipe has a angular acceleration wrt to the truck that is CCW. But the mass center of the pipe has an acceleration to the right.
@@ColinSelleck But shouldn't the acceleration of the mass centre be consistent with the angular acceleration?
@@younamsayin No. The alpha is wrt to truck and it can be CCW but the mass center can be accelerating to the right. Look at the relative acceleration equation.
@@ColinSelleck i'm pretty confused here... I thought the motion of this truck will cause the ring to move towards the left.
IF we were to use angular velocity instead of angular acceleration, and velocity instead of acceleration,
then in this scenario, would the angular velocity become CCW and the velocity to the left?
@@younamsayin The pipe does move to the left wrt to the truck. But the absolute acceleration of the ring is to the right because the truck is accelerating.
At 12:32 it should be mass = 200 and in the second step a negative sign is missing, should be -0.4T
You are correct. Thanks!
@@ColinSelleck Thank you for the amazing videos.
Thank you so much for making and uploading these videos, they have been extremely helpful.
I am having difficulty understanding why in he firs example, you used the inner radius instead of that of he body.
I use the inner radius because the wheel is slipping at the contact point with the ground. You can visualize the cable being wrapped around the inner radius as the wheel moves to the right and the instantaneous center of zero velocity is at point A..
OK! Awesome thank you :)
12:37 You substituted the mass in as 20 instead of 200.
You are correct.
sorry please help, am confused. at 10:00 , y acceleration = 0.4 alpha ??? we have to write that only when there is a fixed point
In 8:32 the spool problem, does anyone know why it moves to the right? I think it should move to the left because the moment is counterclockwise. Thanks
It can't move to the left because of the presence of the cord. The cord is going to wrap itself around the spool and the spool will slip at point B.
At 7:57 you tell us about the acceleration of the mass centre being consistent with the angular acceleration. Can you please elaborate on this, I'm not sure exactly sure about what this means. Thanks
This statement is only true for frictional rolling problems that I had just gone over in the video. If there is no slip (and a homogeneous cylinder so the mass center is at the center of the cylinder), then Ag = r times alpha. If you assume alpha is CCW, then Ag must be assumed to the left and vice versa.
Colin Selleck Thank you so much. If it's not a rolling problem, then can I relaxed with assumed directions of alpha and acceleration of the mass centre?
That is correct.
Colin Selleck thank you. Is there a patron or something that I can contribute to? I would like to support you if there is a way
No patron. I do it to help my fellow (future) engineers.
I love!
Thanks!
how do you know which way the frictional force goes? is it not always away from the movement?
Yes, frictional forces are directed opposite of motion (or impending motion).
First thank you for the video. If the spool in your first example is 1962 [N] / 9.81 = 200 [Kg], Not 20 [Kg] there for your solution is incorrect. Unless I missed something.
Thanks Moshe. You are correct. Moment of Inertia should be 200(0.3)^2.
Hello guys. I attempted fundamental problem 17.13 but saw that the mass of rod wasn't accounted for in the sum of forces in the y-direction, is there some fundamental concept I'm missing or is it just an error on their side(highly unlikely)?
This confused me too at first until I realized the drawing is a plan view (from the top), not a side view. So gravity does not play a role. And since the plane is smooth, neither does friction.
How does a CCW moment cause the wheel to move to the right?
Because of the cable attached at B. The wheel will slip on the ground and the cable will be wrapped around the cylinder.
At 5:37 you tell us not to use F=(mu)*N and the in you first example you at 9:40 you do use 0,1*N …? How does that work?
AT 5:37 I didn't say not to use F=(mu)N. I said first assume no slip (so F is not equal to (mu)N) and solve for the frictional force. Then compare the friction with (mu)N. If it is greater, then use F = (mu)N.