Good morning. Bobby, please read the problem and Billy, please translate.
♪ Flipping Physics ♪ A cart of mass m sub c on a frictionless
surface is attached via a string over a pulley to a hanging mass m sub h. The frictionless pulley
has radius capital R, mass m sub p and rotational inertia about its axis of capital I sub p
which is equal to x times the mass of the pulley times the radius of the pulley squared.
Determine the acceleration of the system. Our knowns are the mass of the pulley,
the hanging mass, the mass of the cart, the radius of the pulley capital R,
the coefficient of friction equals zero because the surface is frictionless. The rotational
inertia of the pulley equals x times the mass of the pulley times the radius of the pulley squared.
And acceleration equals question mark. I don’t get it.
What? What is x?
Oh, most of the moments of inertia we have worked with equal a fraction times the mass of the object
times its radius squared. Like, about its center of mass, for the rotational inertia of a solid
disk, x equals one half, for a thin, hollow sphere x equals two thirds, for a solid sphere x equals
two fifths, for a hoop x equals one. I get it. Thanks.
How do we know we are solving for linear acceleration and not angular acceleration?
I think the general assumption is that, unless the question specifically asks
for the angular quantity, we are talking about the linear quantity, right mr.p?
Correct Bo, that is a good general assumption. Bo, please identify all the forces acting on
the hanging mass, the pulley, and the cart. Okay. That is going to be a lot of forces.
(yep) Starting with the hanging mass there is a
force of gravity acting down on the hanging mass. A force of tension acts up on the hanging mass.
An equal magnitude force of tension acts down on the other end of that string right where
the string comes off the pulley. That is also the force of tension from the hanging mass. Moving to
the cart there is a force of gravity acting down on the cart. There is a force normal acting upward
on the cart. A force of tension acts to the left on the cart. An equal magnitude force of tension
acts to the right on the other end of that string that is also the force of tension from the cart.
(The pulley has mass, right Bo?) The pulley has mass so there is a force of gravity
acting down on the pulley. And therefore, there must also be a force acting upward on the pulley
to prevent it from accelerating downward, let’s call that a force normal acting up on the pulley.
I thought the two forces of tension acting on either side of a pulley had the same magnitude.
Uh that is only true when the pulley is frictionless and massless. Because this pulley
has mass, it has rotational inertia, and resists angular acceleration, which makes it so the two
forces of tension which act on either side of the pulley do not have to be equal in magnitude.
Thanks. Alright, I do want to point
out that the force normal acting on the pulley does not quite act straight up. Because of the
two forces of tension which act on the pulley act down and to the right, the force normal has
a magnitude which is slightly larger than the force of gravity acting on the pulley and the
force normal acts up and slightly to the left. Oh yeah. Because the net force acting on the
pulley equals zero. Okay.
Now, we know net torque equals the derivative of angular momentum with respect to time,
therefore, if we sum the torques acting on the entire system, the hanging mass, the pulley,
and the cart, with an axis of rotation at the center of mass of the pulley, that will
equal the derivative of the total angular momentum of the system with respect to time.
Woah. The whole system? Yeah.
Do the cart and the hanging mass even have angular momentum about the axis of rotation of the pulley?
Yeah, remember the angular momentum of a particle equals the cross product of the object’s position
and momentum? Right…
Let’s determine the angular momenta of all three objects, the pulley, the cart, and the hanging
mass. Bobby, please start by finding the angular momentum of the pulley about its axis of rotation.
Sure. The angular momentum of a rigid body like the pulley equals rotational inertia
times angular velocity. Substitute in x times mass of the pulley times radius of the pulley squared
for rotational inertia. And we know tangential velocity equals radius times angular velocity,
so the angular velocity of the pulley equals the tangential velocity at the outer edge of the
pulley where the string is, divided by the radius of the pulley, and we can substitute that in.
Radius cancels out and we get that the angular momentum of the pulley about its axis of rotation
equals x times the mass of the pulley times the radius of the pulley times the tangential
velocity of the outer edge of the pulley. Thank you Bobby. Billy, please determine
the angular momentum of the cart about the axis of rotation of the pulley.
Absolutely. The angular momentum of a particle equals the cross product of its r position
and momentum which equals the distance from the axis of rotation to the cart, r sub c, times the
mass of the cart times the speed of the cart times the sine of the angle between the direction of
r and the velocity of the cart. But the r position of the cart and the angle for the
cart change as the cart moves toward the pulley…? We did an entire lesson dedicated to this.
Oh yeah. The sine of theta equals opposite over hypotenuse. So, for the cart it equals
the radius of the pulley over the r position of the cart. That means the radius of the pulley,
which is a constant, equals r position of the cart times the sine of theta for the cart
and we can substitute that back into the equation and get that the angular momentum of the cart
equals the radius of the pulley times the mass of the cart times the speed of the cart.
Actually, we can do that exact same thing for the hanging mass and get that the angular
momentum of the hanging mass equals the radius of the pulley times the mass of the
hanging mass times the speed of the hanging mass. And all of those speeds have the same magnitude.
Which speeds? The tangential
speed of the rim of the pulley, the speed of the hanging mass, and the speed of the cart.
Oh cool! Then we can label them all just v to represent the speed of the system.
Thanks. Now we can determine the angular momentum of the whole system;
it equals the sum of the angular momenta of all three objects. And we can substitute in the
equations for angular momenta y’all just solved for. And we can factor out the radius of the
pulley times the speed of the system. Remember, the whole reason we did this is because the net
torque acting on the whole system with an axis of rotation at the center of the pulley equals the
derivative of the total angular momentum of the system with respect to time. So, now we can sum
the torques acting on the whole system. One for every force that acts on any part of the system.
Notice, I have added a torque here caused by every single force in our free body diagrams, however,
I have not yet identified if they are positive or negative. Bobby, that, I am going to let you do.
I get to identify if the torques caused by each of those forces are positive or
negative …? Limber up!
Oh yeah. The right-hand rule! Don’t be too cool.
Definitely not. I would not deign to be.
Okay. From left to right, for the direction of the torque caused by the force of gravity
acting on the hanging mass well, we start the right-hand rule with our right hand at the axis
of rotation of the pulley, point our fingers along the r position vector, and then curl our
fingers in the direction of the force of gravity acting on the hanging mass and we get
out of the screen which is positive. Screen?
(Actually Bobby, could you please define the positive direction?)
Sure. Because the cart moves left, the hanging mass moves down, and the pulley rotates
counterclockwise from our vantage point, let’s define all of those as positive directions. We
show that by adding a quarter circle with an arrow and a plus sign to our diagram. Is that good mr.p?
(Yep. Thanks Bobby. Eight more torque directions to go…)
For the direction of the torque due to the force of tension acting on the hanging mass
we again start at the axis of rotation of the pulley,
point our fingers along the r position vector, and curl our fingers in the direction of the force of
tension acting on the hanging mass and our right thumb points into the screen which is negative.
For the torque caused by the equal magnitude force acting on the other end of that string we get
a positive torque. For the torque caused by the force of gravity acting on the
pulley the right-hand rule does not work because the force acts at the axis of rotation which makes
the r distance zero which makes that torque zero and zero does not have a direction.
The same is true for the torque caused by the force normal acting on the pulley,
it equals zero. Next, we have the direction of the torque caused by the force of tension
acting on the pulley caused by the cart. Starting with our fingers at the axis of rotation pointing
along r, curling in the direction of the force of tension. Thumb points into the screen which is
negative. The torque caused by the force of tension acting on the cart is
positive. For the torque caused by the force of gravity acting on the cart, the torque is
negative. And the torque caused by the force normal acting on the cart is positive. And,
I think those are the directions of all 9 torques. Nice!
Yes. Well done Bobby. Now, there are a bunch of torque pairs here which cancel one another out.
One example is the two torques caused by the forces of tension from the cart.
The two forces of tension from the cart are equal in magnitude and the two torques are opposite in
direction and, because the radius of the pulley equals the r vector for the cart times the sine
of the angle for the cart, the two torques caused by the forces of tension from the cart are equal
and opposite and cancel one another out. The same is true for the two torques caused by the force
of tension from the hanging mass. The two torques are equal and opposite and cancel one another out.
And then there are the torques caused by the force normal acting on the cart and the force
of gravity acting on the cart. The acceleration of the cart in the y-direction is zero,
therefore, the net force in the y-direction on the cart is zero, and the force of gravity acting on
the cart and the force normal acting on the cart are of equal magnitude. And because both torques
have the same magnitude lever arm, the minimum distance from the axis of rotation to the line
of action of the force, those two torques are also equal and opposite, and cancel one another out.
In other words, the only torque which remains is the torque caused by the force of gravity
acting on the hanging mass. (Wow. Uh. Woah.)
Exactly. Bo, please take it from here.
Take what from here? (Solve for the linear acceleration of the system.)
Oh. Right. That’s what we’re doing. Okay. We can substitute r position times force of
gravity times sine theta for the torque caused by the force of gravity on the hanging mass.
We can substitute in the equation we got a long time ago for the angular momentum of the system.
We already know r position of the hanging mass times the sine of the angle for the hanging mass
equals the radius of the pulley, so substitute that into the torque equation and force of gravity
equals mass times acceleration due to gravity. On the right-hand side of the equation actually, the
only variable that changes as a function of time is the speed of the system, so we can take all the
other variables out from the derivative. And … Everybody brought radius to the party!
And the derivative of the magnitude of velocity is the magnitude of the acceleration. The
acceleration of the system equals the hanging mass times acceleration due to gravity all divided
by the quantity x times the mass of the pulley plus the hanging mass plus the mass of the cart.
Very nice everybody. We have solved for the linear acceleration of the system.
It is important to realize that, for the pulley, this is the linear acceleration of the rim
of the pulley, at any radius smaller than capital R, the radius of the pulley,
the linear acceleration of that part of the pulley is smaller than what we just solved for.
(mr.p…?) Yes Billy? We could have solved this by summing the torques
on just the pulley around the axis of rotation of the pulley. Solved that in a similar fashion to
what we just did. And then summed the forces on just the hanging mass in the positive direction
and solved for the force of tension on the hanging mass. And then summed the forces on
just the cart in the positive direction and solved for the force of tension on the cart.
And then substituted both force of tension equations back into the original net torque
equation and solved for acceleration. That gives us the same linear acceleration, right mr.p?
That is correct Billy. (mr.p…?) Yes Bo?
We could have used conservation of mechanical energy instead, right? I mean, if we set the
initial point where everything starts, set the final point after the cart and hanging mass have
moved a distance y, and set the horizontal zero line at the final height of the hanging mass. For
conservation of mechanical energy well, the cart and the pulley have no change in gravitational
potential energy, so we do not need to add them to the equation. Since everything starts at rest,
and the hanging mass is the only object whose gravitational potential energy changes,
that is the only mechanical energy we need to consider at the beginning. At the final point,
the system has rotational kinetic energy of the pulley, and linear kinetic energies of the
hanging mass and the cart. The initial height of the cart is y. The angular velocity final
of the pulley equals the linear velocity final of the system over the radius of the pulley.
We can solve for the square of the final velocity of the system. Because the net force on the system
is constant, the acceleration of the system is constant, so we can use the uniformly
accelerated motion equation velocity final squared equals velocity initial squared plus two times
acceleration times displacement. Substitute in the equation for velocity final squared,
initial velocity is zero, and the displacement of the system is y. And we get the same linear
acceleration of the system we solved for before. That solution works too, right mr.p?
Yes, Bo, that is also a valid solution. There are often multiple ways to solve physics
(mr.p…?) problems. Yes Bobby? We need to test to see if the
physics works, right? Yeah! Yeah.
Sure. We can do that. I measured the mass of the pulley to be 3.62 grams, the hanging mass
is 25.0 grams, the mass of the cart is 271 grams, and x for this pulley equals 0.91.
Plugging all those numbers into the equation for linear acceleration gives us 0.819353 or 0.82
meters per second squared with 2 sig figs. So, 0.82 meters per second squared is the acceleration
we should expect to measure. This is a graph of the measured magnitude of the acceleration of the
cart, and therefore the system, as a function of time. The average of the accelerations once the
cart reaches its full acceleration is 0.790752 or roughly 0.79 meters per second squared.
Using our relative error equation gives us a relative error of roughly
negative 3.5 percent. The most likely culprit for our observed acceleration being slightly less than
our predicted acceleration is friction between the cart and the track, and in the axle of the pulley,
which we neglected. However, I would call that a small relative error. I think that means
The physics works! The physics works! Uh huh, uh huh. The Physics Works!
Thank you very much for learning with me today, I enjoy learning with you.