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Direct Analogies Between Linear Translational and Rotational Motion

This text explores the direct analogies between linear translational and rotational motion in terms of quantity and principle, discussing key concepts like position, velocity, acceleration, inertia, momentum, and kinetic energy. It also delves into the application of rotational energy considerations in practical scenarios like pitching baseballs and rolling objects. The text provides insights before an introductory physics class, offering detailed explanations and examples to aid understanding.

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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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