1/15/16Oregon State University PH 212, Class 61 Here are some of the direct analogies between (linear) translational and rotational motion: Quantity or PrincipleLinear Rotation Displacement x  Velocity v  Acceleration a  Inertia (resistance to mass (m) moment of acceleration) inertia (I) Momentum P = mv L = I  Momentum rate of changedP/dt = F net dL/dt =  net Stated as Newton’s 2 nd Law:  F = ma  = I 

1/15/16Oregon State University PH 212, Class 62 What is a torque? It is a “twisting” force that causes angular acceleration in a rigid body—usually caused by a linear force acting at some point on the body, causing it to rotate about some axis. If the body is not constrained (“pinned”), the rotation axis will be the body’s center of mass (c.m.). Torque is a vector (r x F), positive when its twisting effect is counter-clockwise; negative if clockwise. An easy way to compute this is  = (F · sin  )d where d is the distance from the axis of rotation to the point where F is acting on the body, and  is the angle between F and d. Torque has units of force·distance (N·m).

You are using a wrench to tighten a lug- nut on a car tire. Which of these four arrangements shown is most effective? A. 1 B. 2 C. 3 D. 4 E. All the same 1/15/163Oregon State University PH 212, Class 6

A straight, rigid, mass-less rod of length 1.00 m connects the point masses shown. Where must you locate a fulcrum (support point) so that the object balances (does not tip) with gravity? 1/15/164Oregon State University PH 212, Class 6 1.x = m 2.x = m 3.x = m 4.x = m 5.None of the above kg30.0 kg

1/15/16Oregon State University PH 212, Class 65 Notice that in locating the fulcrum at the center of mass, we effectively balanced the torques acting on the rod. That is, we prevented the stick from accelerating rotationally—i.e. in the angular sense. Indeed, Newton’s First Law must include rotational motion, too: A rigid body is in total mechanical equilibrium only when it is not accelerating along any translational axis; and when it is not accelerating around any rotational axis. In other words, it’s in total mechanical equilibrium when:  F = 0 and  = 0

1/15/16Oregon State University PH 212, Class 66 And notice how summing both torques and forces works together. Consider, for example, item 3d in the Prep 3 set… Follow-up: What force (magnitude and direction) is the fulcrum exerting on the board so that the board remains at rest?

1/15/16Oregon State University PH 212, Class 67 And of course, the two parts of Newton’s 2 nd Law addresses all cases in general—when a rigid body is not necessarily in total mechanical equilibrium:  F = ma and  = I  As always, notice how summing both torques and forces works together. Consider, for example, item 4a in the Prep 3 set…