1 / 12

Problem 31

Problem 31. Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg. Problem 40. m each blade = 160 kg . Moment of inertia I ? Starts from rest, torque τ to get ω = 5 rev/s ( 10 π rad/s ) in t = 8 s ?. Example 8-11. M = 4 kg, F T = 15 N

peregrine
Télécharger la présentation

Problem 31

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Problem 40 m each blade = 160 kg. Moment of inertia I? Starts from rest, torque τto get ω = 5 rev/s (10πrad/s) in t = 8 s?

  3. Example 8-11 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

  4. Example 8-12 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 

  5. Translation-Rotation Analogues & Connections Translation Rotation 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)

  6. Section 8-7: 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)

  7. Translation-Rotation Analogues & Connections Translation Rotation 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)

  8. Sect. 8-7: 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

  9. Example 8-13 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

  10. Conceptual Example 8-14: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)

  11. Friction: Necessary for objects to roll without slipping. Example: work done by friction hasn’t been included (used KE +PE =const). WHY? Because Ffr Motion  Ffrdoes no work!

More Related