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R OTATIONAL W ORK AND P OWER  Recall that work W is given by a force F times a distance s: Topic 2.3 Extended A – Rotational work and kinetic energy.

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Presentation on theme: "R OTATIONAL W ORK AND P OWER  Recall that work W is given by a force F times a distance s: Topic 2.3 Extended A – Rotational work and kinetic energy."— Presentation transcript:

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2 R OTATIONAL W ORK AND P OWER  Recall that work W is given by a force F times a distance s: Topic 2.3 Extended A – Rotational work and kinetic energy W = Fs  If the force is tangent to a rotating object, s = r  so that W = Fs W = Fr  W =  Rotational Work  Recall that power P is the amount of work done in time t: P = W/t.  For force, this translates to P = WtWt Fs t P = Fv  For torque, this translates to P = WtWt  t P =  Rotational Power

3 R OTATIONAL K INETIC E NERGY AND THE W ORK- E NERGY T HEOREM Topic 2.3 Extended A – Rotational work and kinetic energy  The formula for rotational work, can be manipulated as follows: W = , W =  Given W = I  substitution of  = I  2W = I(2  ) Why? 2W = I(  2 -  0 2 )  2 =  0 2 + 2  W = I  2 - I  0 2 1212 1212 Why? W = K - K 0 The Work-Energy theorem K = I  2 1212 Rotational Kinetic Energy Work-Kinetic Energy Theorem

4 C ONSERVATION OF E NERGY Topic 2.3 Extended A – Rotational work and kinetic energy  Suppose a hoop, a disk, and a solid sphere, all of the same radius R and mass M are released to roll from the same place at the same time on a ramp. Which one will reach the bottom first? M,R,IM,R,I h v 0 = 0 v  Let’s solve this problem using energy: K – K 0 + U – U 0 = 0 U 0 = K U 0 = K trans + K rot Mgh = + Iω2Iω2 1212 Mv cm 2 1212 v cm = Rω v cm 2 R 2 ω 2 = Mgh = + v cm 2 1212 Mv cm 2 1212 IR2IR2 v cm = 2Mgh M + IR2IR2 the ramp roll race

5 C ONSERVATION OF E NERGY Topic 2.3 Extended A – Rotational work and kinetic energy  Suppose a hoop, a disk, and a solid sphere, all of the same radius R and mass M are released to roll from the same place at the same time on a ramp. Which one will reach the bottom first?  Now we can finally answer the question: v cm = 2Mgh M + MR2R2MR2R2 I hoop = MR 2 I disk = MR 2 1212 I sphere = MR 2 2525 v cm = 2Mgh M + MR22R2MR22R2 v cm = 2Mgh M + 2MR25R22MR25R2 v cm = gh v cm = gh 4343 v cm = gh 10 7  Now we can finally answer the question:  The sphere has the greatest velocity – therefore it wins the race… followed by the disk, then the hoop. Don’t forget: The work done by friction is ZERO in rolling motion, because no slippage occurs.

6 S O M ANY G &% D @#&!D F ORMULAS... Topic 2.3 Extended A – Rotational work and kinetic energy But Brain, how are we going to remember all these formulas? Easy, Pinky. Just remember the correspondences between linear and rotational variables! F ↔ τ m ↔ I a ↔ α v ↔ ω s ↔ θ Translation between linear and rotational formulas


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