Great episode! I just realized that this is why the moment of inertiavis always smallest when pivoting at its center of mass! It also predicts that when comparing moments of inertia about different points, it is purely the distance from the COM that determines their relative values. All the AP teachers at my APSI for APCM had a hard time justifying why they chose a particular answer on an APCM MC question; they knew it was due to the distance from the COM, but noone could state why.
I think he set it to zero because the x and y coordinates of our center of mass are set at the origin of our coordinate plane. And the equation he used i think was a derived equation from our center of mass equation x_cm = net_i ( m_i * x_i ) / m_tot and same thing for the y i think. that is the same thing as saying x_cm = 1/m_tot * int(x dm) i think. then he set the integral to zero because he multiplied both sides by m_tot and it yielded zero I think
"why is X(cm) and Y(cm) = zero?" Because r is defined as the distance from the origin (and the center of mass) to dm. So, the center of mass is where r = 0 and therefore both x_cm and y_cm = 0. "why are they equal to 1/m intergral x dm ?" See: www.flippingphysics.com/center-mass-integral.html
I have a question what does quantities which are result of cross product have direction , perpendicular to the plane of the productant like torque what does it mean
In pure math, a cross product of two vectors both in distance units, corresponds to the area of the parallelogram formed by the two vectors when located head-to-head, and the remaining two sides completed per the symmetry in the definition of a parallelogram. Area vectors are perpendicular to the surface with the area, and which perpendicular direction is determined by which vector is CCW from the other. The fact that angular kinematics and kinetics quantities are perpendicular to the plane of rotation is just a matter of convention. It doesn't mean anything physical is happening in that direction, but rather because there is no preference between the two directions in the plane of rotation, there is no preference to either one being the vector direction. The axis of rotation and right hand rule, is the convention in physics for how to keep track of these quantities. We could've just as easily assigned it according to a left hand rule, so the choice was ultimately arbitrary. But in order to communicate with others, you learn to follow the standard convention.
Thank you Sir - I wonder The theorem can use in case of "A New Parallel is located outside the object - such that simple pendulum Given : pendulum ball with some radius r and is suspened with a rope (a rope is small mass ) So the center of mass of this system is center of mass of the ball. When we release the ball , then moment of inertia of system is I (cm) + (M+R)^2 [ if the ball is sphere : 2/5MR^2 + (M+R)^2 ] , isn't it ? And The other question - When we consider conservation of energy of The pendulum Should we care about size of the ball pendulum ? ( I think We should because while the ball is moving down with radius R each point mass on the ball have different velocity - mean that v = omega R Why don't we use 1/2 Iw^2 instead of 1/2mv^2 ) Thank you [ Apologize for my english 'I am not english speaker😄 ]
You've used both the capital R and lowercase r in your example, to stand for different measurements, and then lost track of this distinction. Given the radius of the ball as lowercase r, and length of the pendulum from the pivot to the ball's center is capital R. To translate the axis of rotation to a parallel axis at the end of the string, you add M*dcm^2 to the original rotational inertia, where dcm is the distance between the center of mass and the new axis of rotation. In your case, this distance is R. So in your example, the new rotational inertia is: I = M*R^2 + 2/5*M*r^2 This is its rotational inertia regardless of whether it is at rest, or whether you release it and it starts rotating. The moment of inertia is independent of the kinematics of this system. It only depends on mass, shape, axis of rotation and distribution of mass. This is a fixed value as long as the pendulum moves as a rigid body. A ball on a string moves approximately as a rigid body when swinging as a pendulum, but for a better example that really is a rigid body, consider replacing the string with a rigid rod of negligible mass. "Should we care about size of the ball pendulum?" In the formula I provided, you do account for the size of the ball. In the limit as r/R approaches zero, you notice that it becomes simply I=M*R^2, which is the moment of inertia of a point mass around an axis at distance R from it. "v = omega R Why don't we use 1/2 Iw^2 instead of 1/2mv^2 ) " Because angular velocity (omega) is the same value for every point on a rigid body, by virtue of it being a rigid body. Linear velocity by contrast, is different for every single point. Using 1/2*I*omega^2 for kinetic energy is a shortcut to allow you to calculate rotational kinetic energy without needing to use Calculus to keep track of how velocity varies throughout the object. In concept, you can set up an integral to calculate linear velocity of every point on the body, and add up the kinetic energy of every point mass that comprises the body. However, it is much simpler to take credit for the fact that omega is the same throughout the rigid body, and use the property of moment of inertia instead. v=omega*r is only valid at the point that is distance r away from the axis of rotation. If r is a constant in your problem, instead of a variable coordinate, then not every point is a distance r from the axis of rotation. If by contrast, r is a variable coordinate, then every point on the body has a different value of r.
Wow fantastic job explaining the proof. You sir are a hero!
Great episode! I just realized that this is why the moment of inertiavis always smallest when pivoting at its center of mass! It also predicts that when comparing moments of inertia about different points, it is purely the distance from the COM that determines their relative values. All the AP teachers at my APSI for APCM had a hard time justifying why they chose a particular answer on an APCM MC question; they knew it was due to the distance from the COM, but noone could state why.
Good point. Thanks!
eeh you time traveller why does your comments are 1 month old
you are such a gem
please keep uploading
Amazing explanation ever, Thank you so much!!!!
I love the way you explain the theorem. thank you so much.
You deserve more views
you are so telented at teaching!
Talented😂
A rigorous one.. loved it ❤️:))
Great video. Thank you!
6:25 Mr. P where did you come up with that equation? Moreover, I couldn't catch why the integral is to be zero. Please elucidate on that!
I think he set it to zero because the x and y coordinates of our center of mass are set at the origin of our coordinate plane. And the equation he used i think was a derived equation from our center of mass equation x_cm = net_i ( m_i * x_i ) / m_tot and same thing for the y i think. that is the same thing as saying x_cm = 1/m_tot * int(x dm) i think. then he set the integral to zero because he multiplied both sides by m_tot and it yielded zero I think
damn man, amazing derivation! thanks a lot!
Sir could you please make a video on angular momentum in case of rotation about fixed axis
Thank you
HELPPPPP why is X(cm) and Y(cm) = zero?????
and why are they equal to 1/m intergral x dm ????? 6:57
"why is X(cm) and Y(cm) = zero?"
Because r is defined as the distance from the origin (and the center of mass) to dm. So, the center of mass is where r = 0 and therefore both x_cm and y_cm = 0.
"why are they equal to 1/m intergral x dm ?"
See: www.flippingphysics.com/center-mass-integral.html
@@FlippingPhysics Oh thanks to all four copies of you :)
I have a question
what does quantities which are result of cross product have direction , perpendicular to the plane of the productant
like torque what does it mean
In pure math, a cross product of two vectors both in distance units, corresponds to the area of the parallelogram formed by the two vectors when located head-to-head, and the remaining two sides completed per the symmetry in the definition of a parallelogram. Area vectors are perpendicular to the surface with the area, and which perpendicular direction is determined by which vector is CCW from the other.
The fact that angular kinematics and kinetics quantities are perpendicular to the plane of rotation is just a matter of convention. It doesn't mean anything physical is happening in that direction, but rather because there is no preference between the two directions in the plane of rotation, there is no preference to either one being the vector direction. The axis of rotation and right hand rule, is the convention in physics for how to keep track of these quantities. We could've just as easily assigned it according to a left hand rule, so the choice was ultimately arbitrary. But in order to communicate with others, you learn to follow the standard convention.
Looks good!
Thanks!
I love you man
Absolutely stunning explanation 😍
Glad you think so!
nice
Great work
Thank you Sir - I wonder The theorem can use in case of "A New Parallel is located outside the object - such that simple pendulum
Given : pendulum ball with some radius r and is suspened with a rope (a rope is small mass ) So the center of mass of this system is center of mass of the ball. When we release the ball , then moment of inertia of system is I (cm) + (M+R)^2
[ if the ball is sphere : 2/5MR^2 + (M+R)^2 ] , isn't it ?
And The other question - When we consider conservation of energy of The pendulum Should we care about size of the ball pendulum ?
( I think We should because while the ball is moving down with radius R each point mass on the ball have different velocity - mean that
v = omega R Why don't we use 1/2 Iw^2 instead of 1/2mv^2 )
Thank you [ Apologize for my english 'I am not english speaker😄 ]
You've used both the capital R and lowercase r in your example, to stand for different measurements, and then lost track of this distinction.
Given the radius of the ball as lowercase r, and length of the pendulum from the pivot to the ball's center is capital R. To translate the axis of rotation to a parallel axis at the end of the string, you add M*dcm^2 to the original rotational inertia, where dcm is the distance between the center of mass and the new axis of rotation. In your case, this distance is R.
So in your example, the new rotational inertia is:
I = M*R^2 + 2/5*M*r^2
This is its rotational inertia regardless of whether it is at rest, or whether you release it and it starts rotating. The moment of inertia is independent of the kinematics of this system. It only depends on mass, shape, axis of rotation and distribution of mass. This is a fixed value as long as the pendulum moves as a rigid body. A ball on a string moves approximately as a rigid body when swinging as a pendulum, but for a better example that really is a rigid body, consider replacing the string with a rigid rod of negligible mass.
"Should we care about size of the ball pendulum?"
In the formula I provided, you do account for the size of the ball. In the limit as r/R approaches zero, you notice that it becomes simply I=M*R^2, which is the moment of inertia of a point mass around an axis at distance R from it.
"v = omega R Why don't we use 1/2 Iw^2 instead of 1/2mv^2 ) "
Because angular velocity (omega) is the same value for every point on a rigid body, by virtue of it being a rigid body. Linear velocity by contrast, is different for every single point. Using 1/2*I*omega^2 for kinetic energy is a shortcut to allow you to calculate rotational kinetic energy without needing to use Calculus to keep track of how velocity varies throughout the object. In concept, you can set up an integral to calculate linear velocity of every point on the body, and add up the kinetic energy of every point mass that comprises the body. However, it is much simpler to take credit for the fact that omega is the same throughout the rigid body, and use the property of moment of inertia instead.
v=omega*r is only valid at the point that is distance r away from the axis of rotation. If r is a constant in your problem, instead of a variable coordinate, then not every point is a distance r from the axis of rotation. If by contrast, r is a variable coordinate, then every point on the body has a different value of r.
gou are awesome
thank gou
@@FlippingPhysics 😜