Torque | 8.01 Classical Mechanics,
case of a massless rod with two equal masses at both ends. and I rotate about this
axis perpendicular to the blackboard.. There will be no force on that pivot point
about which the two rotate., but if I take a torque relative to point p, any force
through point p has no effect Because the position vector is zero.. If there is no
force at all, then the torque relative toany point must be zero.. The spin angular
momentum is an intrinsic property of a spinning object. it is the same relative to
any point that you choose. even if I choose a point here in space, you can prove
that it is still the same angular momentum. there are a huge number of
applications, and in the next four or five lectures we will go through many of
them.. I do this bang very short hit. and now. The question is what will this
object do. what does your instinct? Tell you. what will it rotate. and if it
rotates, about which point. What must always hold. for the center of mass.?
center of mass behaves like a point source f equals ma and the velocity of the
center of mass only depends on I and on the mass of the object. the larger I, the
larger I, the larger the velocity, the lower the velocity. now you have to choose
an origin, because you 're dealing with torques, and so you would probably choose C
as your origin. and you can clearly see that there is a torque relative to point
c.. if you hit right in the middle of the center of mass, you don't expect it to
rotate at all. so clearly there must be a torque about the center. and so now, all
you would have to do is to relate that change in angular momentum with this
distance D and with i. I. and you will see that if D becomes larger, that indeed
the omega of center of. mass, the angular velocity will increase. it 's very
sensitive. we were able to calculate the period of oscillation of a pendulum. that
was relatively easy. but now suppose you have a ruler like this, and you have here
a little pinhole, and you're going to oscillate it like this. that looks like a
mathematical headache. and yet, with the knowledge that we have now, this can be
solved quite easily..
even a system as bizarre as this hula hoop, which oscillates about that pin, can
now be calculated easily. so let 's first do the.... We first take the ruler. this
is a ruler. it rotates about the pin perpendicular to the blackboard.. for sure,
there will be forces through that point P. theta in time must be some maximum angle
times. Cosine of omega T plus phi.. This omega hasnothing to do with that omega.
This is angular velocity, and this is angular frequency.. This will never change.
that 's related to the period of oscillation. this, by the way, is independent of
the mass of the object.. This is an incredibly complex system that is going to
rotate about a pin, which is offset from the center. in our case, a b equals 40
centimeters, and it is an exact meter stick. we could. Meter stick. we could be off
by something like maybe two millimeters. so it 's 40 centimeters plus or minus 0. 2
centimeters. now, once you have done this with your ruler, you can now apply this
knowledge on a way more interesting system..
case of a massless rod with two equal masses at both ends. and I rotate about this
axis perpendicular to the blackboard.. There will be no force on that pivot point
about which the two rotate., but if I take a torque relative to point p, any force
through point p has no effect Because the position vector is zero.. If there is no
force at all, then the torque relative toany point must be zero.. The spin angular
momentum is an intrinsic property of a spinning object. it is the same relative to
any point that you choose. even if I choose a point here in space, you can prove
that it is still the same angular momentum. there are a huge number of
applications, and in the next four or five lectures we will go through many of
them.. I do this bang very short hit. and now. The question is what will this
object do. what does your instinct? Tell you. what will it rotate. and if it
rotates, about which point. What must always hold. for the center of mass.?
center of mass behaves like a point source f equals ma and the velocity of the
center of mass only depends on I and on the mass of the object. the larger I, the
larger I, the larger the velocity, the lower the velocity. now you have to choose
an origin, because you 're dealing with torques, and so you would probably choose C
as your origin. and you can clearly see that there is a torque relative to point
c.. if you hit right in the middle of the center of mass, you don't expect it to
rotate at all. so clearly there must be a torque about the center. and so now, all
you would have to do is to relate that change in angular momentum with this
distance D and with i. I. and you will see that if D becomes larger, that indeed
the omega of center of. mass, the angular velocity will increase. it 's very
sensitive. we were able to calculate the period of oscillation of a pendulum. that
was relatively easy. but now suppose you have a ruler like this, and you have here
a little pinhole, and you're going to oscillate it like this. that looks like a
mathematical headache. and yet, with the knowledge that we have now, this can be
solved quite easily..
even a system as bizarre as this hula hoop, which oscillates about that pin, can
now be calculated easily. so let 's first do the.... We first take the ruler. this
is a ruler. it rotates about the pin perpendicular to the blackboard.. for sure,
there will be forces through that point P. theta in time must be some maximum angle
times. Cosine of omega T plus phi.. This omega hasnothing to do with that omega.
This is angular velocity, and this is angular frequency.. This will never change.
that 's related to the period of oscillation. this, by the way, is independent of
the mass of the object.. This is an incredibly complex system that is going to
rotate about a pin, which is offset from the center. in our case, a b equals 40
centimeters, and it is an exact meter stick. we could. Meter stick. we could be off
by something like maybe two millimeters. so it 's 40 centimeters plus or minus 0. 2
centimeters. now, once you have done this with your ruler, you can now apply this
knowledge on a way more interesting system..
