Work and Power for Rotation Work = Fd = FRq t = FR q F s s = Rq Work = tq Power = = Workt tq t w = q t Power = t w Power = Torque x average angular velocity

Example 1: The rotating disk has a radius of 40 cm and a mass of 6 kg Example 1: The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s. q F F=W s s = 20 m 2 kg 6 kg Work = tq = FR q sR q = = = 50 rad 20 m 0.4 m F = mg = (2 kg)(9.8 m/s2); F = 19.6 N Work = 392 J Work = (19.6 N)(0.4 m)(50 rad) Power = = Workt 392 J 4s Power = 98 W

The Work-Energy Theorem Recall for linear motion that the work done is equal to the change in linear kinetic energy: Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy:

Two Kinds of Kinetic Energy  v R P Kinetic Energy of Translation: K = ½mv2 Kinetic Energy of Rotation: K = ½I2 Total Kinetic Energy of a Rolling Object:

Translation or Rotation? If you are to solve for a linear parameter, you must convert all angular terms to linear terms: If you are to solve for an angular parameter, you must convert all linear terms to angular terms:

Total energy: E = ½mv2 + ½Iw2 Example 2: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v. Compare the kinetic energies. w v Two kinds of energy: KT = ½mv2 Kr = ½Iw2 w = vR Total energy: E = ½mv2 + ½Iw2 Disk: E = ¾mv2 Hoop: E = mv2

Conservation of Energy The total energy is still conserved for systems in rotation and translation. However, rotation must now be considered. Begin: (U + Kt + KR)o = End: (U + Kt + KR)f Height? Rotation? velocity? mgho ½Iwo2 ½mvo2 = mghf ½Iwf2 ½mvf2 Height? Rotation? velocity?

= v = 8.85 m/s mgho mghf ½Iwo2 ½Iwf2 ½mvo2 ½mvf2 Example 3: Find the velocity of the 2-kg mass just before it strikes the floor. h = 10 m 6 kg 2 kg R = 50 cm mgho ½Iwo2 ½mvo2 = mghf ½Iwf2 ½mvf2 2.5v2 = 196 m2/s2 v = 8.85 m/s

mgho = ½mv2 + ½mv2; mgho = mv2 Example 4: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m? mgho = ½mv2 + ½Iw2 Hoop: I = mR2 20 m mgho = ½mv2 + ½mv2; mgho = mv2 Hoop: v = 14 m/s Disk: I = ½mR2; mgho = ½mv2 + ½Iw2 v = 16.2 m/s

Angular Momentum Defined Consider a particle m moving with velocity v in a circle of radius r. m2 m3 m4 m m1 axis w v = wr Object rotating at constant w. Define angular momentum L: L = mvr Substituting v= wr, gives: L = m(wr) r = mr2w Since I = Smr2, we have: L = Iw For extended rotating body: L = (Smr2) w Angular Momentum

Example Rank the following from largest to smallest angular momentum.

answer

m = 4 kg L = 2 m Example 5: Find the angular momentum of a thin 4-kg rod of length 2 m if it rotates about its midpoint at a speed of 300 rpm. For rod: I = mL2 = (4 kg)(2 m)2 1 12 I = 1.33 kg m2 L = Iw = (1.33 kg m2)(31.4 rad/s)2 L = 1315 kg m2/s

Impulse and Momentum Recall for linear motion the linear impulse is equal to the change in linear momentum: Using angular analogies, we find angular impulse to be equal to the change in angular momentum:

Impulse = change in angular momentum Example 6: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? R 2 kg w F wo = 0 rad/s R = 0.40 m F = 200 N D t = 0.002 s I = mR2 = (2 kg)(0.4 m)2 I = 0.32 kg m2 Applied torque t = FR Impulse = change in angular momentum t Dt = Iwf - Iwo FR Dt = Iwf wf = 0.5 rad/s

Conservation of Momentum If no net external torques act on a system then the system’s angular momentum, L, remains constant. Speed and radius can change just as long as angular momentum is constant. Ifwf = Iowo Ifwf - Iowo = t Dt Io = 2 kg m2; wo = 600 rpm If = 6 kg m2; wo = ? wf = 200 rpm

Example The figure below shows two masses held together by a thread on a rod that is rotating about its center with angular velocity, ω. If the thread breaks, what happens to the system's (a) angular momentum and (b) angular speed. (Increase, decrease or remains the same)

Examples of the Conservation of Angular Momentum Natalia Kanounnikova World Record Spin - 308 RPM

Diving 18 meter high dive Amazing Fouette Turns on Pointe Playground Physics

Summary – Rotational Analogies Quantity Linear Rotational Displacement Displacement x Radians  Inertia Mass (kg) I (kgm2) Force Newtons N Torque N·m Velocity v “ m/s ”  Rad/s Acceleration a “ m/s2 ”  Rad/s2 Momentum mv (kg m/s) I (kgm2rad/s)

Analogous Formulas F = ma  = I K = ½mv2 K = ½I2 Work = Fx Work = tq Linear Motion Rotational Motion F = ma  = I K = ½mv2 K = ½I2 Work = Fx Work = tq Power = Fv Power = I Fx = ½mvf2 - ½mvo2  = ½If2 - ½Io2

= Summary of Formulas: I = SmR2 mgho mghf Height? ½Iwo2 ½Iwf2 ½mvo2 = mghf ½Iwf2 ½mvf2 Height? Rotation? velocity?

CONCLUSION: Angular Motion