Newton’s Second Law for Rotational Motion F = ma τ = Fr Substituting for F τ = (ma) r Remember aT = rα Substituting for acceleration τ = (m rα) r τ = (mr2) α Moment of Inertia, I = mr2 Walk students through the derivation steps. Ask students for the fundamental difference between tangential and rotational acceleration.

Moment of Inertia Moment of Inertia, I = mr2 Resistance of an object to rotation (degree of laziness) Different for different shapes For rotating objects, Newton’s Law becomes τ=Iα α must be in rad/s2 How is rotational inertia compared to linear inertia? Direct students to page 291 to see a chart for rotational inertia of standard objects.

Example Three particles of masses of 2.3kg, 3.2kg and 1.5kg are connected by thin rods of negligible mass, so that they lie at the vertices of 3-4-5 right triangle in the x-y plane. y (m) =4.5N 3 c Make sure students draw the picture. The order of the masses are sequential, so m1 = 2.3, m2 = 3.2, and m3 = 1.5. X (m) 4

Find the rotational inertia about each of the three axes perpendicular to the x-y plane and passing through one of the particles. A force of magnitude 4.5N is applied to m2 in the xy plane and makes an angles of 300 with the horizontal. Find the angular acceleration about oz axis. Find the rotational inertial about an axis perpendicular to the xy plane and passing through the center of mass of the system 1) Students should work to find a solution to each part, in their notebooks. Work time given is approximately 7 minutes.

Solution: through y (m) =4.5N 3 c 4 X (m) Make sure students are using the right equation to find rotational inertia. Why didn’t students need to use one of the equations from page 291? =4.5N 3 c 4 X (m)

Solution: Ask students why the angle of 30 degrees was used. Make sure students were still comfortable applying the center of mass equation.

Solution: 1) Why was it necessary to find each r?

The parallel-axis theorem The result of the previous sample problem leads us to an important general result, the parallel-axis theorem: the rotational inertia of any body about an arbitrary axis equals the rotational inertial about a parallel axis through the center of mass plus the total mass times the squared distance between two axes . Students should be familiar with the term ‘parallel-axis theorem’ but do not need to memorize the conceptual definition. Explain that Icm = moment of inertia through the center of mass, M = total mass of the object, h = distance between the axis of rotation and the axis through the center of mass.

Students do not need to memorize these. Ask students to rationalize one or 2 of these equations (ie. Where does the equation originate from for a spherical shell?) This is the same chart as is found on page 291.

Sample A student opens a 12kg door by applying a constant force of 40N at a perpendicular distance of .9m from the hinges. If the door is 2m in height and 1m wide, what is the magnitude of its angular acceleration? τ = I α α = rF / (1/3)ML2 α = ((3)(.9)(40)) / ((12)(12)) = 9 rad/s2 1) Give students about 4 minutes to arrive at the answer before revealing it.