2. Translational-Rotational Analogues & Connections Continue! Translation Rotation Displacementx θ Velocityvω Accelerationaα Force (Torque)Fτ Massm? CONNECTIONS v = rω a tan = rα a R = (v 2 /r) = ω 2 r τ = r  F

3. Section 8-5: Rotational Dynamics Goal: Newton’s 2 nd Law for rotational motion. We’ve just seen: The angular acceleration is  the net torque or α  τ net = ∑τ = sum of torques Analogous to Newton’s 2 nd Law: a  F net = ∑F = sum of forces –Recall, Newton’s 2 nd Law: ∑F = ma. a  α, F  τ Question: What plays the role of mass m for rotational problems?

4. Simplest Possible Case A mass m moving in a circle of radius r, one force F TANGENTIAL to the circle τ = rF Newton’s 2 nd Law + relation (a = rα) between tangential & angular accelerations  F = ma = mrα So τ = mr 2 α  Newton’s 2 nd Law for Rotations Proportionality constant between τ & α is mr 2 (point mass only!)

5.  For one particle, moving in a circle, Newton’s 2 nd Law for rotational motion is: τ = mr 2 α Extrapolate to a rigid body, made of many particles. Newton’s 2 nd Law, rotational motion becomes ∑τ = ∑(mr 2 )α –Since α is the same for all points in the body, it comes out of the sum.  Write ∑τ = Iα, where the proportionality constant I = ∑(mr 2 ) = m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 + …  Moment of Inertia of the body sum over all point masses in the body  

6.  Newton’s 2 nd Law, rotational motion is: The net torque = (moment of inertia)  (angular acceleration) Moment of Inertia of the body: Moment of Inertia I = A measure of the rotational inertia of the body. Analogous to the mass m = A measure of translational inertia of the body.

7. Translational-Rotational Analogues & Connections Translation Rotation Displacementx θ Velocityvω Accelerationaα Force (Torque)Fτ Mass (moment of inertia) mI CONNECTIONS v = rω, a tan = rα, a R = (v 2 /r) = ω 2 r τ = r  F, I = ∑(mr 2 ) NEWTON’S 2 nd LAW: ∑F = ma; ∑τ = Iα

8. Obtained for simple shaped bodies using calculus. Useful for problem solving!

9. Moment of Inertia: I = ∑(mr 2 ) Depends on mass m AND on the distribution of mass (r 2 ) in the object! I disk > I rod, even if their masses are the same!!!   Disk or thin cylinder   Long, small radius cylinder

10. Example 8-10 I = ∑(mr 2 ) a) I = (5)(2) 2 +(7)(2) 2 = 48 kg m 2 b) I = (5)(0.5) 2 + (7)(4.5) 2 = 143 kg m 2 NOTE! I depends on the rotation axis and on the mass distribution