1 considering the setup in figure 2 please use eqs 3 5 to...

Considering the setup in Figure 2, please use Eqs. (3)–(5) to show that the torque exerted on the rotating object is τ = m_h r (a − α r). The total rotational kinetic energy of the system is K_total = 1/2 (I + m_h r^2) ω^2. In this lab, we will use these relations to compute example quantities from the measurements. For example, we will use the measured m_h, r, and α to compute the torque τ, and then the moment of inertia as I = τ/α. In the real experiment with the setup of Figure 2, initially the disk is at rest and the hanging mass is placed 0.500 m above the floor. Then the hanging mass falls down until it finally touches the floor. Measurements show that the hanging mass m_h = 0.0251 kg, the spindle radius r = 0.115 m, the angular acceleration α = 3.551 rad/s^2, and the disk's final angular velocity ω_f = 11.29 rad/s. Please compute the quantities below: - Hanging mass's acceleration a - Hanging mass's initial potential energy U - String tension T - Hanging mass's final velocity v_h - Torque exerted on disk τ - Hanging mass's final kinetic energy K_h - Disk's moment of inertia I - Disk's final rotational kinetic energy K_d A disk with moment of inertia 9.15 × 10^−3 kg·m^2 initially rotates about its center at angular velocity 5.32 rad/s. A non-rotating ring with moment of inertia 4.86 × 10^−3 kg·m^2 is dropped onto the disk. Finally, the two objects rotate at the same angular velocity ω_f about the same axis. There is no external torque acting on the system during the collision. Please compute the system's quantities below: - Initial angular momentum L_i - Final angular velocity ω_f - Initial rotational kinetic energy K_i - Final rotational kinetic energy K_f

Transcript

00:01 Here i'll be analyzing a measurement of moment of inertia that one can do in the laboratory.

00:08 And there's quite a bit of pieces to it in terms of mechanics, both dynamics as in duton's second law as well as energy.

00:20 But the idea is that there is a small mass that hangs down over a pulley.

00:26 So, yeah, that should be attached to something that pulley.

00:30 Like at the table.

00:34 That pulley is lightweight and frictionless, we hope.

00:40 The weight is tied by a string or rope around a spindle on an axle, so to speak, of radius, little r.

00:52 And then there is a large, usually iron object disk that sits on top of the turntable that the spindle is attached to.

01:07 And as the mass drops, as the hanging mass, little m sub h is a hanging mass, as that mass drops, the plate starts to rotate.

01:19 And how this all works together is it's probably best to start off with the idea of conservation of energy is one idea that you need to analyze the situation.

01:34 The idea is that the little mass starts off initially, some height above the floor, and it may be all the way up at the table, but we'll just show that it starts off at some h above the floor.

01:52 So it starts off with initial potential energy, and it converts part of that potential energy into kinetic energy.

02:10 So we have u -initial, and we'll assume that k -initial is zero.

02:20 The sum of kinetic plus potential energy has got to be conserved.

02:25 So u -final will consider the mass all the way at the floor plus k -final.

02:34 The initial potential energy is mgh, the hanging mass times gravity, times the height, that it begins to drop.

02:46 The kinetic energy final is a sum of the kinetic energy of the little mass, hanging mass, plus the kinetic energy of the rotational table.

03:02 We'll call that the r, the rotational table.

03:07 We can add that up.

03:09 The small mass has one half, m, h, b squared.

03:13 The rotation table has one half times the moment of inertia times omega and we should put little subscripts final for each of those.

03:26 And it turns out that because of the linear connection between the table and the hanging mass, what we have is omega -final times r the radius of the spindle is equal to be final because of the linkage between the hanging mass and the rotating table.

03:49 So we can put all this together.

03:52 We don't need to, but we can put it all together and say that the mgh initial.

04:06 So we'll call this equation one.

04:08 M .g .h is equal to 1 1�m .h.

04:18 Omega -squared final times r squared plus one -half moment of inertia, omega -final squared.

04:29 So what you can measure in this system, usually the hanging mass, the final rotational angular velocity, the radius of the spindle, and the initial height through which this is going to drop, this hanging mass is going to drop.

04:47 And putting that all together, you can solve for the moment of inertia.

04:54 Okay, and finally, for some other things, if you can get all that, that's usually what the experiment involves, but you can also analyze the dynamics.

05:08 So in our next step, we're going to look at the sum of f equals ma for the system consisting of the two objects.

05:26 So newton's second law for the system.

05:31 So it's a system of two connected objects.

05:36 And we will go ahead and draw the forces on the hanging mass.

05:44 So there are two of them.

05:45 There's the weight downwards, mh times g.

05:53 And there is the tension in the string upwards.

05:58 Let's see, we'll use purple for the tension in the string.

06:04 Now, supposedly the pulley is frictionless and lightweight enough so that the tension will transmit all the way up to the spindle.

06:17 And we'll have the tension pulling the opposite way on the spindle, exact identical tension.

06:30 And we'll call that big t.

06:34 So there are two objects, we'll call them object a.

06:40 And object b.

06:44 And for object a, our sum of forces looks like mhg minus the tension has got to equal mh times a.

07:00 And for object b, the plate being pulled by the rope, we have basically, it's not quite f equals ma, but it's the equivalent.

07:14 The sum of torques equals i times alpha.

07:20 Torque is a product of force times moment arm, and so it's a little bit hard to see, but if we were looking down on that spindle, what we would see is the spindle looking down on it, has a rope wrapped around it.

07:39 Okay, yeah, we almost need a different color.

07:43 And then the rope is coming off at a right angle with the tension at the radius r.

07:50 So for object b, we have some of torques equals i alpha.

07:57 There's only one torque, which is tension times r is equal to i times alpha.

08:05 And again, there is a linear relationship because of the linkage between the hanging mass and the rotating table.

08:14 We have a is equal to alpha times r.

08:18 There is a constraint in there.

08:23 So we can put all this together.

08:26 We can substitute alpha in the acceleration in equation a.

08:41 Okay, so we can do that.

08:45 Give myself some space to write.

08:48 So a becomes, if we substitute alpha r, and if we substitute tension, tension equals i alpha divided by r.

09:09 If we put that all back into equation a with the substitutions, what we get is mh times g minus i alpha over r is equal to, let's see, is that what we want to do? let me stop and recall what we want to replace with what? there are lots of ways to go with this.

09:54 Yes, we want to solve for the torque on the rotating object, not the tension, although it's nice to get the tension involved in that as well.

10:03 So we want to replace tension with torque.

10:21 Let's see.

10:22 So a bit of a step there is equal to tension times r, or tension equals tension.

10:50 Times r.

10:54 So tension is torque over r.

10:58 Now we can substitute that back into a and get mhg minus torque over r is equal to mh alpha times r...