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 acceleration) mass (m) moment of inertia (I) Momentum P = mv L = I d. P/dt = Fnet d. L/dt = net F = ma = I Momentum rate of change Stated as Newton’s 2 nd Law: Work F • Ds • D Kinetic energy (1/2)mv 2 (1/2)I 2 1/25/19 OSU PH 212, Before Class 9 1

When KT and KR may both be useful The total kinetic energy of an object that is rotating around any fixed axis is most easily computed as “pure rotational energy: ” Ktotal = KR. fixed-axis = (1/2)Ifixed-axisw 2 Now note: Ifixed-axis = Icm + Md 2 (parallel axis theorem) So: Ktotal = (1/2)Icmw 2 + (1/2)Md 2 w 2 But: d is the speed, vc. m. , of the center of mass as it rotates around the fixed axis. So: Ktotal = (1/2)Icmw 2 + (1/2)Mvcm 2 = KT. cm + KR. cm In general (fixed axis or free rotation): Ktotal = KT. cm + KR. cm 1/25/19 OSU PH 212, Before Class 9 2

When KT and KR may both be useful Option 1: Ktotal = KR. fixed-axis = (1/2)Ifixed-axisw 2 Option 2: Ktotal = KR. cm + KT. cm = (1/2)Icmw 2 + (1/2)Mvcm 2 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? ) 1/25/19 OSU PH 212, Before Class 9 3

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. 1/25/19 OSU PH 212, Before Class 9 4

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 mass moves 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! 1/25/19 OSU PH 212, Before Class 9 5

Conclusion: When an object is rolling without slipping, you can express its total kinetic energy completely in terms of either or vcm. That is: (1/2)mvcm 2 = (1/2)m(r )2 = (1/2)m. R 2 2 Or: (1/2)I 2 = (1/2)I(vcm/R)2 = (1/2)Ivcm 2/R 2 Example: A hollow spherical shell of mass M and I = (2/3)MR 2, rolling without slipping at speed vcm, has this total kinetic energy: Ktotal = KT. cm + KR. cm = (1/2)Mvcm 2 + (1/2)Ivcm 2/R 2 = (1/2)Mvcm 2 + (1/2)(2/3)MR 2 vcm 2/R 2 = (5/6)Mvcm 2 1/25/19 OSU PH 212, Before Class 9 6