Fall2023 Moment of Inertia Lab Online-1

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Physics

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Apr 3, 2024

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docx

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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 = , 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 = 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
Note: a hollow cylinder is also known as a ring. The difference between the hollow cylinder and the thin- walled cylinder is that for the thin-walled cylinder the outer and inner radii are so close in value that they can be treated as being of the same value, while for the regular hollow cylinder that isn’t true. One way to measure the moment of inertia of a rigid body experimentally is to attach it to a fixed pivot point allowing for rotation in the horizontal plane, apply a known constant torque to it, then measure its tangential acceleration, and finally do a little algebra. One step up that can be used to do this is a two pulley system. One horizontal pulley centered about the pivot point, and another vertical pulley. A string will be wrapped around the first pulley, and then strung over the vertical pulley with a known mass m h attached to it. The mass will be released, causing a constant tension in the string, which will in turn cause a constant torque about the first pulley, finally resulting in the system experiencing a constant acceleration. In such a setup we start with the basic equation relating the applied torque to the moment of inertia. τ = So the moment of inertia will be the torque divided by the angular acceleration of the system. I = τ α 3
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
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