**PLEASE NOTE** At 12:30 I forgot to square the Vt. I did remember to square the R which is why I could cancel them out. The numerical answer is correct because I did the work before had. -1 for me!** (This edit was picked up by a viewer and I thank you for letting me know!)
thank you so much for amazing video... but i have a question... for the second problem isn't like the initial high height of the ball on the incline h = radius + height of the incline initially?
Not exactly, Gravity is the Force that makes it roll. Fg parallel is a “component” of the weight that causes the block to accelerate. Components often times are not asked to be drawn on Free Body Diagrams The force of the normal is an actual force that coincidently is equal to the other component of the weight Hope that helps
So essentially, rolling without slipping makes it so that the total energy of the final will be the sum of the tangential kinetic energy and the rotational energy? I am a bit confused about when I should be equating the gravitation potential energy or any other potential energy to both the tangential and rotational kinetic energy.
It is called translational kinetic energy, by the way. Not tangential kinetic energy. Any time both energies could come in to play, the net work done on the object has to equal the change in both kinds of kinetic energy. An object in pure translation with all of its velocities being uniformly the same, has exclusively translational kinetic energy. An object rotating around its center of mass has exclusively rotational kinetic energy. An object in general that moves as a rigid body could have both kinds of energy coexisting simultaneously. Rolling without slipping means that there is the kinematic constraint that the linear velocity is locally zero at the point of contact, so that there is no skidding between the surfaces in contact. This means that the linear velocity of its center of rotation (v) will be related to its angular velocity (omega) via v=omega*r. Think of a gear rolling down a cog railway. The gear teeth cannot skip a cog. A condition of rolling without slipping is the same thing, except instead of gear teeth establishing the constraint, the force of static friction (aka traction) establishes the constraint. So putting together: v=omega*r, KEt = 1/2*m*v^2,, KEr = 1/2*I*omega^2, and KEnet = KEt + KEr This takes into consideration, the constraint of rolling without slipping and conservation of energy, in order to enable you to solve for v, based on the work done accelerating it from rest to obtain KEnet.
This explanation of why the static friction force does no work is popular on the internet, but I'm just not buying it. The point of contact is the base of the wheel, but the bonds in the wheel transmit the force to the rest of the wheel. Translationally, we apply Newton's 2nd law to the center of mass. The static friction force opposes the motion of the center of mass and therefore does negative translational work on the wheel. The static friction force also produces a torque that does positive rotational work. Presumably these two cancel out (didn't work it out yet) so the net work is 0. In your example, the ball needs some force to the right.
since the normal force is not coming from the center as you said, does the normal produce a torque? Ugh, this sounds like a dumb question for some reason!!
The Normal Force goes through the CENTER of the Circle, or disk, therefore it can NOT produce a Torque. Any force that acts Through the Center of Rotation contributes ZERO force to the Torque, therefore it does not contribute to the Motion.
**PLEASE NOTE** At 12:30 I forgot to square the Vt. I did remember to square the R which is why I could cancel them out. The numerical answer is correct because I did the work before had. -1 for me!** (This edit was picked up by a viewer and I thank you for letting me know!)
Im self studying and these videos give me so much confidence that I am learning thank you
You got this!
thank you so much for amazing video... but i have a question... for the second problem isn't like the initial high height of the ball on the incline h = radius + height of the incline initially?
You can make the height whatever you want. It’s a variable
u are explaination god, thank you
Thank you so much for watching!
Thank you so much for this amazing explanation!
You're very welcome!
wow great video, just watched all energy videos and im rdy
Excellent. Almost there
Great stuff. Thank you so much.
You're very welcome!
With the forces on the ball, what about Fg Parallel? That's the main force making it roll in the first place, right?
Not exactly, Gravity is the Force that makes it roll. Fg parallel is a “component” of the weight that causes the block to accelerate. Components often times are not asked to be drawn on Free Body Diagrams
The force of the normal is an actual force that coincidently is equal to the other component of the weight
Hope that helps
Math And Physics Tutor OHH THANK U
So essentially, rolling without slipping makes it so that the total energy of the final will be the sum of the tangential kinetic energy and the rotational energy? I am a bit confused about when I should be equating the gravitation potential energy or any other potential energy to both the tangential and rotational kinetic energy.
rolling without slipping means that friction doesn't do any work
It is called translational kinetic energy, by the way. Not tangential kinetic energy. Any time both energies could come in to play, the net work done on the object has to equal the change in both kinds of kinetic energy. An object in pure translation with all of its velocities being uniformly the same, has exclusively translational kinetic energy. An object rotating around its center of mass has exclusively rotational kinetic energy. An object in general that moves as a rigid body could have both kinds of energy coexisting simultaneously.
Rolling without slipping means that there is the kinematic constraint that the linear velocity is locally zero at the point of contact, so that there is no skidding between the surfaces in contact. This means that the linear velocity of its center of rotation (v) will be related to its angular velocity (omega) via v=omega*r. Think of a gear rolling down a cog railway. The gear teeth cannot skip a cog. A condition of rolling without slipping is the same thing, except instead of gear teeth establishing the constraint, the force of static friction (aka traction) establishes the constraint.
So putting together:
v=omega*r,
KEt = 1/2*m*v^2,,
KEr = 1/2*I*omega^2,
and
KEnet = KEt + KEr
This takes into consideration, the constraint of rolling without slipping and conservation of energy, in order to enable you to solve for v, based on the work done accelerating it from rest to obtain KEnet.
great vidoe sir ,looking for more
Cheers
is there a likelihood of "rolling with slipping" on this year's exam?
your guess is as good as mine. I dont make the test.. I take the test
This explanation of why the static friction force does no work is popular on the internet, but I'm just not buying it. The point of contact is the base of the wheel, but the bonds in the wheel transmit the force to the rest of the wheel. Translationally, we apply Newton's 2nd law to the center of mass. The static friction force opposes the motion of the center of mass and therefore does negative translational work on the wheel. The static friction force also produces a torque that does positive rotational work. Presumably these two cancel out (didn't work it out yet) so the net work is 0. In your example, the ball needs some force to the right.
Oh ok
Very nice!
Yes it is
since the normal force is not coming from the center as you said, does the normal produce a torque? Ugh, this sounds like a dumb question for some reason!!
No. Torque is produced by a force perpendicular to the “r”
The Normal Force goes through the CENTER of the Circle, or disk, therefore it can NOT produce a Torque. Any force that acts Through the Center of Rotation contributes ZERO force to the Torque, therefore it does not contribute to the Motion.