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Direct analogies between (linear) translational and rotational motion:

Direct analogies between (linear) translational and rotational motion: Quantity or Principle Linear Rotation Position x  Velocity v  Acceleration a  Inertia (resistance to mass ( m ) moment of

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Direct analogies between (linear) translational and rotational motion:

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  1. Direct analogies between (linear) translational and rotational motion: Quantity or PrincipleLinearRotation Position x Velocity v Acceleration a Inertia (resistance to mass (m) moment of acceleration) inertia (I) Momentum P= mvL= Iw Momentum rate of changedP/dt = FnetdL/dt = net Stated as Newton’s 2nd Law: F = ma = Ia Work F•Dst•Dq Kinetic energy (1/2)mv2 (1/2)I2 Oregon State University PH 212, Class 9

  2. One end of a thin, uniform rod (L = 1.40 m, M = 0.325 kg) is attached to a stationary pivot. The rod is free to rotate about the pivot without friction or air resistance. Initially, the rod is held at rest in the “3 o’clock” position. Then it is released to swing freely. Find its total kinetic energy when it has reached the “6 o’clock” position. [Irod.center = (1/12)ML2Irod.end = (1/3)ML2g = 9.8 m/s2] A.Ktotal = 4.46 J B.Ktotal = 2.23 J C.Ktotal = 0.212 J D. Not enough information. E. None of the above. Oregon State University PH 212, Class 9

  3. One end of a thin, uniform rod (L = 1.40 m, M = 0.325 kg) is attached to a stationary pivot. The rod is free to rotate about the pivot without friction or air resistance. Initially, the rod is held at rest in the “3 o’clock” position. Then it is released to swing freely. Find the speed of its center of mass when it has reached the “6 o’clock” position. [Irod.center = (1/12)ML2Irod.end = (1/3)ML2g = 9.8 m/s2] A.vcm = 3.21 m/s B.vcm = 3.70 m/s C.vcm = 4.58 m/s D. Not enough information. E. None of the above. Oregon State University PH 212, Class 9

  4. When KT and KR may both be useful We have seen that the total kinetic energy of an object that is rotating around anyfixed axis is “pure rotational energy:” Ktotal = KR.fixed-axis = (1/2)Ifixed-axisw2 Now note: Ifixed-axis = Icm+ Md2 (parallel axis theorem) So: Ktotal = (1/2)Icmw2 +(1/2)Md2w2 But: dw is the speed, vc.m., of the center of mass as it rotates around the fixed axis. So: Ktotal= (1/2)Icmw2 + (1/2)Mvcm2 = KT.cm + KR.cm In general(fixed axis or free rotation): Ktotal = KT.cm + KR.cm Oregon State University PH 212, Class 9

  5. When KT and KR may both be useful Option 1:Ktotal = KR.fixed-axis = (1/2)Ifixed-axisw2 Option 2:Ktotal= KR.cm + KT.cm = (1/2)Icmw2 + (1/2)Mvcm2 Option 1 is valid only for an object rotating around a fixed axis, but that includes an axis that is only momentarily fixed (i.e. its v = 0 for just an instant). Option 2 is valid for either an object rotating around a fixed axis or a freely rotating object (i.e. rotating around its c.m.). After class 9 notes will go through a couple of examples to demonstrate each option. Note: When an object is rotating around a moving axis that is not the center of mass, Ktotal is not generally a constant value; it is changing in time, because the axis pin is doing work on the object. (So, why doesn’t an unmoving axis pin do work on an object?) Oregon State University PH 212, Class 9

  6. Rotational kinetic energy of a rolling object Rolling: A common example of translation and rotation at the same time. ASSUMPTION: no slipping – the center of mass moves one circumference forward—much like the string on a pulley rim or the chain on a bike sprocket. vcm = Rω Notice that the point of contact on the ground is stationary! Oregon State University PH 212, Class 9

  7. When a solid, uniform disk is rolling at speed vcm along a surface (without slipping), what percentage of its total kinetic energy is due to rotation around its center of mass? [Idisk.center = (1/2)MR2] A. 66.7% B. 50.0% C. 33.3% D. Not enough information E. None of the above. Oregon State University PH 212, Class 9

  8. A bowling ball rolls without slipping, first along a level track, then up a ramp onto another level section of the track, gaining 0.340 m in altitude. If its translational speed along the lower track level was 3.15 m/s, find its translational speed at the upper track level. [Isolid.sphere.center = (2/5)MR2g = 9.80 m/s2] A. 2.58 m/s B. 2.27 m/s C. 0.569 m/s D. Not enough information E. None of the above. Oregon State University PH 212, Class 9

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