Fall2023 Moment of Inertia Lab Online-1

Rotational Inertia Lab Online Purpose The purpose of this exercise is to examine the moment of inertia of both a ring and disk, and to experimentally confirm that the moment of inertia of an object is a function of both its mass and how that mass is spatially distributed. Theory Let us assume there is a mass m, initially at rest, that is attached to one end of a massless rod of length r, and the other end of the rod is attached to a frictionless pivot that is free to rotate 360 o . If one were to apply a net force F to the mass it would induce rotational motion about the pivot point, but only the component of the applied for force that is tangential (at a right angle) to the length r would contribute to the change in motion of the mass. By Newton’s Second Law we know the tangential force would be related to the tangential acceleration by the following equation: F T = ma T The torque acting about the point of rotation that is associated with the tangential component of the applied force would be given by: τ = F T r = mr a T We also know that the angular acceleration that the mass is experiencing is related to its tangential acceleration by a T = rα , so we can insert that into the equation, giving us: τ = m r 2 α What would happen if you would apply a force to a rigid body (solid object) that is fixed in location, but free to rotate about an axis? Now a rigid body is just a collection of the point masses that make the total mass M of the rigid body. Each of those point masses m i will have its own direct line distance r i from the axis of rotation and itself. So for each point mass we could go through the exact same argument as we just went through to arrive at a similar equation; τ i = m i r i 2 α 1

The infinitesimal amount of toque τ i for each point mass is the product of the term m i r i 2 and α the angular acceleration of the rigid body. The angular acceleration does not have a subscript on it because as a rigid body all the point masses rotate together, so all the point masses have the exact same angular acceleration. To find the total toque acting on the entire rigid body you simply sum up all the torques acting on all the point masses. τ = ∑ i n ( m i r i 2 ) α The summation term is given a name. It is called the moment of inertia of the rigid body, and its SI units are kg·m 2 . I = ∑ i n ( m i r i 2 ) This allows us to rewrite our equation as τ = Iα The moment of inertia is a quantification of how difficult (or easy) it is to get an object to change its current state of rotational motion about a particular axis or rotation. The value of the summation of the term m i r i 2 will depend on the total mass of the rigid body, its shape, and the axis of rotation that is picked. However, the summation will always have the following basic algebraic form: I = ∑ i n ( m i r i 2 ) = CM r 2 Here M is the total mass of the object, r is the ‘radius’ of the object, and C is a coefficient dependent on the shape of the object. A chart of the moment of inertia of some basic geometric shapes with uniform mass is given. 2

From the free body diagram and force summation equations of the system, we see that the tension in the string will be: T = m h g − m h a T = m h ( g − a T ) Inserting this for the tension, as well as the known equations for torque and angular acceleration, gives us: I = r∙m h ( g − a T ) a T r = r 2 ∙m h ( g a T − a T a T ) I = r 2 ∙m h ( g a T − 1 ) This final equation gives the experimental value for the moment of inertia of the rigid body, where here r is the radius of the horizontal pulley the rigid body is attached to. (Please note that in constructing this equation we assumed that the moment of inertia of the two pulleys are so small compared to the moment inertia of the rigid body’s that they were simply ignored.) Setup 1. Go to the following website: https://phet.colorado.edu/en/simulations/torque 2. You should now see the following: 4

3. Click on the Play Button to download the Legacy Java Version, and then open the software when it has completed downloading. Additionally, you may be able to run the Browser-Compatible Version directly from website. This does not work for some computers. 4. You should now see the following: (Ignore the ladybug, and the other bug. They won’t be used in our experiment) 5. Click on the Moment of Inertia tab located near the top left of your screen. Afterwards you should see the following: 5

2. At the center left side of your screen there are the controls to select the physical properties of ‘platform’, which will be the Ring in our experiment. a. Set Outer Radius to 4.0 m, and record this value in the data table for Ring. b. Set Inner Radius to 3.5 m. c. Set Platform Mass to 0.20 Kg, and record this value in the data table for Ring. d. Set Force of Break to 0.0 N 3. Near the top of your screen, and slightly off center to the left you will see the setting for the Applied Torque. Set the Applied Torque to 10 Nm, and record this in your data table for Ring. 4. Right below the Applied Torque settings click on the Go button to start the experiment. 5. Record the values for T applied , I, and α platform in the table for Ring. a. The Moment of Inertia you record in the table will serve as your experimental value of the Moment of Inertia of the Ring. Analysis of Rotational Inertia Lab Online Name: Course/Section: Phy-1611-014 Instructor: Tables (25 points) Outer Radius Inner Radius Mass (kg) Torque(Nm) α (rad/s 2 ) I (kgm 2 ) Disk 4.0m 0.20 10.0 6.25 1.600 Ring 4.0m 3.5m 0.20 10.0 3.54 2.825 7

1. Calculate the experimental value of the moment of inertia of the disk, and show work. Hint: Do not use the experimental moment of inertia given for the hanging mass example. (5 points) 𝐼 = 𝑇 𝑎 𝐼 = 10 6.25 = 1.6 𝑘𝑔𝑚 2 2. Calculate the theoretical value of the moment of inertia of the disk, and show work. Hint: Use the appropriate formula from Table 9.2 for the calculation. (5 points) 𝐼 = 1 2 𝑀 𝑅 2 𝐼 = 1 2 ( 0.2 ⋅ ( 4.0 2 ) ) = 1.6 𝑘𝑔𝑚 2 3. Calculate the experimental value of the moment of inertia of the ring, and show work. Hint: Do not use the experimental moment of inertia given for the hanging mass example. (5 points) 𝐼 = 𝑇 𝑎 𝐼 = 10 3.54 = 2.8248 ≈ 2.825 𝑘𝑔𝑚 2 4. Calculate the theoretical value of the moment of inertia of the ring, and show work. Hint: Use the appropriate formula from Table 9.2 for the calculation. (5 points) 𝐼 = 1 2 𝑀 ( 𝑅 1 2 + 𝑅 2 2 ) 𝐼 = 1 2 ( 0.2 ( ( 4.0 2 ) + ( 3.5 2 ) ) ) = 2.825 𝑘𝑔𝑚 2 5. Using the theoretical value as the accepted value, calculate the % error for the disk, and show work. (5 points) 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟 = | h 𝑡 𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 − exp 𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙𝑣𝑎𝑙𝑢𝑒 h 𝑡 𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 | ⋅ 100 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟 = | 1.6 − 1.6 1.6 | ⋅ 100 = 0% 6. Using the theoretical value as the accepted value, calculate the % error of the ring, and show work. (5 points) 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟 = | h 𝑡 𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 − exp 𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙𝑣𝑎𝑙𝑢𝑒 h 𝑡 𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 | ⋅ 100 8