Rotational Dynamics Continued

Find the moments of inertia about the x & y axes: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg

Problem m each blade = 160 kg. Moment of inertia I? Starts from rest, torque τ to get ω = 5 rev/s (10π rad/s) in t = 8 s?

Example M = 4 kg, FT = 15 N Frictional torque: τfr = 1.1 m N t = 0, ω0 = 0; t = 3 s, ω = 30 rad/s I = ? Calculate the angular acceleration: ω = ω0+ αt, α = (ω - ω0)/t = 10 rad/s2 N’s 2nd Law: ∑τ = Iα FTR - τfr = Iα  I = [(15)(0.33) -1.1]/10 I = 0.385 kg m2

a  Example 8-12 a) α = ? (pulley) b) t = 0, at rest. The same pulley, connected to a bucket of weight mg = 15 N (m = 1.53 kg). M = 4 kg I = 0.385 kg m2; τfr = 1.1 m N a) α = ? (pulley) a = ? (bucket) b) t = 0, at rest. t = 3 s, ω = ? (pulley) v = ? (bucket) a 

Translation-Rotation Analogues & Connections Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law: ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 ? CONNECTIONS v = rω, atan= rα, aR = (v2/r) = ω2 r τ = rF, I = ∑(mr2)

Rotational Kinetic Energy Translational motion (Ch. 6): (KE)trans = (½)mv2 Rigid body rotation, angular velocity ω. Rigid  Every point has the same ω. Body is made of particles, masses m. For each m at a distance r from the rotation axis: v = rω. The Rotational KE is: (KE)rot = ∑[(½)mv2] = (½)∑(mr2ω2) = (½)∑(mr2)ω2 ω2 goes outside the sum, since it’s the same everywhere in the body As we just saw, the moment of inertia, I  ∑(mr2)  (KE)rot = (½)Iω2 (Analogous to (½)mv2)

Translation-Rotation Analogues & Connections Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 CONNECTIONS v = rω, atan= rα, aR = (v2/r) = ω2 r τ = rF, I = ∑(mr2)

Rotational + Translational KE Rigid body rotation: (KE)rot = (½)Iω2 Now, consider a rigid body, mass M, rotating (angular velocity ω) about an axis through the CM. At the same time, the CM is translating with velocity vCM Example, a wheel rolling without friction. For this, we saw earlier that vCM = rω. The KE now has 2 parts: (KE)trans & (KE)rot  Total KE = translational KE + rotational KE KE = (KE)trans + (KE)rot or KE = (½)M(vCM)2 + (½)ICMω2 where: ICM = Moment of inertia about an axis through the CM

Example  y = 0 KE+PE conservation: where KE has 2 parts!! A sphere rolls down an incline (no slipping or sliding). KE+PE conservation: (½)Mv2 + (½)Iω2 + MgH = constant, or (KE)1 +(PE)1 = (KE)2 + (PE)2 where KE has 2 parts!! (KE)trans = (½)Mv2 (KE)rot = (½)Iω2  v = 0, ω = 0 v = ?   y = 0

Conceptual Example: Who Wins the Race? v increases as I decreases! Demonstration! MgH = (½)Mv2 + (½)ICMω2 Gravitational PE is Converted to Translational + Rotational KE! Hoop: ICM = MR2 Cylinder: ICM = (½)MR2 Sphere: ICM = (2/5)MR2 (also, v = ωR)

Friction is necessary for objects to roll without slipping Friction is necessary for objects to roll without slipping. In the example, the work done by friction hasn’t been included (we used KE +PE = const). WHY? Because the friction force Ffr is  to the motion direction so it does no work!