Physics Section 8-5 to 8-6 Rotational Dynamics

Moment of Inertia We saw previously that mass can be defined by an object’s ability to resist motion. This is called inertial mass. The rotational analog of mass is called moment of inertia, or rotational inertia. The symbol is I. The units are kg•m2.

A single revolving point mass, m. Values for I A single revolving point mass, m. m r I = mr2 axis of rotation A set of n point masses. axis of rotation m4 n r1 I = Σmiri2 r4 m1 i r2 r3 m2 m3

Moments for various objects I = ⅟ M(a2 + b2) I = ⅟ ML2 12 12 I = MR2 I = ⅖ MR2 I = ⅓ ML2 I = ⅔ MR2 I = ⅟ Ma2 I = ½ MR2 12

Rotational dynamics α ∝ ∑ τ a ∝ ∑ F m r F = ma F = mrα point mass We know that the angular acceleration is proportional to the net torque applied to a system. m α ∝ ∑ τ  r This is analogous to the statement of Newton’s second law for constant mass: point mass a ∝ ∑ F For a single revolving point mass: F = ma Since aT = rα, we can say: F = mrα

Since the motion of the point mass at any point is perpendicular to the radius vector (θ = 90°), we can find the torque by multiplying both sides by the magnitude of the radius vector: rF = mr2α Torque for a single point mass: τ = mr2α For several point masses the sum is: ∑ τ = (∑mr2)α Notice that both these equations contain the moment of inertia, I, for the system in question. In general then, net torque is given by: ∑ τ = I α

Newton’s Second Law τ replaces F, I replaces m, and α replaces a: Newton’s second law applies to rotating objects. τ replaces F, I replaces m, and α replaces a: τnet = I α

I = ⅟ ML2 I = I = ⅓ ML2 I = I = (1.20 kg) (1.83 m)2 = 0.335 kg•m2 12 1. A Japanese quarterstaff (bō) has a mass of 1.20 kg, and measures 183 cm (length) by 3.00 cm (diameter). (a) Calculate the moment of inertia if twirled while holding the center of the staff. I = ⅟ ML2 12 (1.20 kg) (1.83 m)2 I = = 0.335 kg•m2 12 This assumes a rod a lot longer than its diameter. A real 3-D rod should use the equation: I = 1/12 ML2 + 1/4 MR2. In this problem, this would only add an additional 0.0000675 kg•m2! (b) Calculate the moment of inertia if swung while holding the end of the staff. (1.20 kg) (1.83 m)2 I = ⅓ ML2 I = 3 I = 1.34 kg•m2

I = ½ MR2 I = 0.363 kg•m2 f = (0.5)(15 kg)(0.220 m)2 = α = ∆ω/∆t α = 2. A solid steel wheel has a mass of 15.0 kg and a diameter of 0.440 m. It starts from rest. You want to make it rotate up to 8 rot/s within a time of 15.0 s. (a) What torque must be applied to the outer rim of the wheel? Solve for moment of inertia, I: M = 15.0 kg I = ½ MR2 = (0.5)(15 kg)(0.220 m)2 R = 0.220 m I = 0.363 kg•m2 ω0 = 0 rad/s Solve for angular acceleration, α: f = 8 rot/s α = ∆ω/∆t ω = 2π(8) rad/s α = (50.3 – 0 rad/s) ÷ 15.0 s ω = 50.3 rad/s α = 3.35 rad/s2 t = 15.0 s

τnet = I α τnet = τnet = τnet = rFsinθ τnet = rF τnet Solve for torque: τnet = I α τnet = (0.363 kg•m2)(3.35 rad/s2) τnet = 1.21 Nm (b) If you apply this torque by wrapping a strap around the outside of the wheel, how much force should you exert on the strap? τnet = rFsinθ θ = 90° τnet = rF τnet r 1.21 Nm 0.220 m F = = = 5.50 N