Direct analogies between (linear) translational and rotational motion:

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 OSU PH 212, Before Class 8

2. Rotational Kinetic Energy Suppose you have a solid sphere sitting at rest. It has no kinetic energy at all—no motion. Now push on it with a linear force. It accelerates (in the direction of the force) to a non-zero linear velocity. Now it’s moving—it has kinetic energy: KT = (1/2)mv2 Where did that energy come from? It was the work done by the force: W = Fcos·sThat’s the displacement multiplied by the force that is acting in the direction of that displacement. Now suppose the sphere is again sitting at rest. This time, instead of pushing on it with a force, you twist on it—exert a torque. It accelerates (in the angular direction of the torque) to some non-zero angular velocity. Again, it’s moving, so it must have kinetic energy. How do we measure and describe this rotational kinetic energy, KR —and where did it come from? OSU PH 212, Before Class 8

3. First, realize that KR is not a “new kind of energy.” It’s just S(1/2)mivi2, for each tiny portion of mass, mi, in the sphere. But at every point in a rotating object, vi = riw. So the sum becomes KR = S(1/2)miri2w2 = (1/2)(w2)Smiri2 = (1/2)Iw2 And where did this energy come from? It was the work done by the torque: W = ·D (notice the units here). It’s the angular displacement multiplied by the torque acting in the (rotational) direction of that displacement. So we need to expand our definition of the total mechanical energy, Emech, of an object: Emech = KT + KR + UG + US And the Work-Energy relationship: Wext = Emech Wext = KT + KR + UG + US where Wext now includes all Fextcos·s and all ext·D. OSU PH 212, Before Class 8

4. Direct analogies between (linear) translational and rotational motion: Quantity or PrincipleLinearRotation Displacement 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 8

5. Example: A solid disk, initially at rest—lying flat on a frictionless surface—has a mass of 50 kg and a radius of 2 m. The disk is pinned at one edge (so that it can only rotate horizontally around that pin). A torque of 20 N·m is applied to the disk until it has rotated for 100 radians. What angular velocity does it then have? [Idisk.center = (1/2)MR2] A. = 3.65 rad/s B. = 13.3 rad/s C. = 14.6 rad/s D. Not enough information E. None of the above OSU PH 212, Before Class 8

6. Example: A solid disk, initially at rest—lying flat on a frictionless surface—has a mass of 50 kg and a radius of 2 m. The disk is pinned at one edge (so that it can only rotate horizontally around that pin). A torque of 20 N·m is applied to the disk until it has rotated for 100 radians. What angular velocity does it then have? [Idisk.center = (1/2)MR2] A. = 3.65 rad/s B. = 13.3 rad/s C. = 14.6 rad/s D. Not enough information E. None of the above (Don’t forget to use the Parallel Axis Theorem to find Idisk.edge!) OSU PH 212, Before Class 8

7. 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 8

8. Same situation, different calculation: A solid disk, initially at rest—lying flat on a frictionless surface—has a mass of 50 kg and a radius of 2 m. The disk is pinned at one edge (so that it can only rotate horizontally around that pin). A torque of 20 N·m is applied to the disk until it has rotated for 100 radians. Find the disk’s KT.cm and KR.cm at that time. [Idisk.center = (1/2)MR2] A.KT.cm = 91.3 J KR.cm = 2000 J B.KT.cm = 1333 J KR.cm = 667 J C.KT.cm = 1000 J KR.cm = 1000 J D. Not enough information E. None of the above OSU PH 212, Before Class 8

9. Same situation, different calculation: A solid disk, initially at rest—lying flat on a frictionless surface—has a mass of 50 kg and a radius of 2 m. The disk is pinned at one edge (so that it can only rotate horizontally around that pin). A torque of 20 N·m is applied to the disk until it has rotated for 100 radians. Find the disk’s KT.cm and KR.cm at that time. [Idisk.center = (1/2)MR2] A.KT.cm = 91.3 J KR.cm = 2000 J B.KT.cm = 1333 J KR.cm = 667 J C.KT.cm = 1000 J KR.cm = 1000 J D. Not enough information E. None of the above Notice: The sum of these two must equalKR.pin (= Ktotal = 2000 J). OSU PH 212, Before Class 8