Mr.p: Good morning,
Previously we determined the rotational inertia of the rotational inertia demonstrator from Arbor Scientific.
In order to do so, we measured the force of tension acting on the hanging mass,
which has the same magnitude as the force of tension acting on the pulley.
Let's take a look at how the measured force of tension changes depending on the acceleration of the pulley.
"Flipping Physics" Intro
You can see the measured force of tension decreases as soon as I let go of the pulley
and the hanging mass begins accelerating downwards.
In order to understand why this happens we need, of course, the free body diagrams.
The force of gravity on the pulley acts down at its center,
which is also the location where the upward normal force acts on the pulley.
On both ends of the string are two equal magnitude forces of tension;
one acts down on the pulley, the other acts up on the hanging mass.
And the hanging mass includes both the force sensor and the mass which is hanging from it.
And there is a force of gravity acting down on the center of mass of the hanging mass.
Let's make a counter clockwise, or out of the board, as the positive torque direction.
Bobby, please determine what the force of tension should be before I let go of the pulley.
Bobby: Do we not need the mass of the hanging mass to solve for that?
Mr.p: Oh, I'm sorry, right. Um, the hanging mass in this example is 0.103 kilograms,
which again includes the mass of the force sensor.
Bobby: OK, thanks.
Now that we know that, we can sum the forces in the y-direction acting on the hanging mass.
Force of tension minus the force of gravity equals...
Bo: Actually, we defined counterclockwise, or out of the board, as positive,
therefore down on that side of the pulley is positive and up is negative.
Bobby: Right. Okay, so...
force of gravity minus force of tension equals mass times acceleration in the y-direction.
This is before you let go of the pulley, so the system is at rest and the acceleration the y-direction equals zero.
Therefore the force of tension equals the force of gravity
which equals the hanging mass times acceleration due to gravity or 0.103 times 9.81
which equals 1.01043 or 1.01 newtons.
Mr.p: Thanks, Bobby.
And notice that, before I let go of a pulley,
the averaged measured force of tension does work out to be 1.01 newtons with three significant digits.
Mr.p: Billy, considering we know the radius of the pulley is 0.0202 meters
and the angular acceleration of the pulley is 29.910 radians per second squared,
please solve for the force of tension after I let go of the pulley.
Billy: Actually,
we just use the same equation Bobby used, only the acceleration in the y-direction is no longer is zero.
Therefore the force of tension equals the force of gravity
minus mass times acceleration in the y-direction.
We can substitute in the equation for force of gravity and factor out mass.
We know the acceleration in the y-direction of the hanging mass
is the same as the tangential acceleration
of the rim of the pulley, which equals the radius of the pulley times angular acceleration.
So the force of tension equals to mass times the quantity acceleration due to gravity
minus the radius of the pulley times angular acceleration
or 0.103 times the quantity 9.81 minus 0.0202 times 29.910, which is 0.948199, or 0.948 newtons.
Mr.p: Thanks, Billy.
Our average force of tension while the system is accelerating measures out to be 0.955 newtons
which I would consider to be very close to the predicted force of tension of 0.948 newtons.
So, please realize the force of tension in the string depends on the acceleration of the hanging mass
or the angular acceleration of the pulley. As soon as the system is allowed to accelerate,
the force of tension in the string decreases.
Thank you very much for learning with me today, I enjoyed learning with you.