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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 OSU PH 212, Before Class 9

  2. When KT and KR may both be useful The total kinetic energy of an object that is rotating around anyfixed axis is most easily computed as “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 OSU PH 212, Before Class 9

  3. 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?) OSU PH 212, Before Class 9

  4. A simple application of rotational energy considerations: A professional pitcher and catcher are testing a new design for a baseball. The mass and radius of the new ball (B) are the same as the current ball (A), but A is a solid sphere, and B has a hollow center. Q: How could this skilled pitcher and catcher duo tell the two baseballs apart? A: The hollow-centered ball would be more difficult to put “spin” on (i.e. throw a curve ball), because it has a larger moment of inertia—that’s a larger resistance to angular acceleration. So the pitcher would have to expend more energy (do more work) with every pitch. OSU PH 212, Before Class 9

  5. Rotational kinetic energy of a rolling object Rolling: A common example of translation and rotation at the same time. ASSUMPTION: no slipping – so the center of massmoves one circumference forward—much like the string on a pulley rim or the chain on a bike sprocket—when the rim rotates by one revolution. vcm = Rω Notice that the point of contact on the ground is stationary! OSU PH 212, Before Class 9

  6. Conclusion: When an object is rolling without slipping, you can express its total kinetic energy completely in terms of eitherw or vcm. That is: (1/2)mvcm2 = (1/2)m(rw)2 = (1/2)mR2w2 Or: (1/2)Iw2 = (1/2)I(vcm/R)2 = (1/2)Ivcm2/R2 Example: A hollow spherical shell of mass M and I = (2/3)MR2, rolling without slipping at speed vcm, has this total kinetic energy: Ktotal = KT.cm + KR.cm = (1/2)Mvcm2 + (1/2)Ivcm2/R2 = (1/2)Mvcm2 + (1/2)(2/3)MR2vcm2/R2 = (5/6)Mvcm2 OSU PH 212, Before Class 9

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