11/7 Rotational Energy

1. 11/7 Rotational Energy • Today: Rotational Energy Examples • HW: 11/6 HW Handout “Rotational Energy” due Wednesday 11/13 • Exam 3: Thursday, 11/14 • Monday holiday-no lab next week!

2. KER Rotational Kinetic Energy • Just another bucket KE = 1/2mv2 Ring has same m and same v so KER,Ring = KET,Block Disk has same m but mass near center is slower so less KER here KER,Disk = 1/2KET,Block Ring of mass m rotating with velocity v Disk of mass m rotating with velocity v Block of mass m moving with velocity v

3. Proportional Reasoning Rolling without slipping means that vR = vT Find total KE of ring and disk KET = 24J KET = 24J KET = 24J KER = 12J KER = 24J Total KE = 24J Total KE = 48J Total KE = 36J

4. Proportional Reasoning Rolling without slipping means that vR = vT Now all have Total KE = 72J. Find the velocity if the mass is 2kg. KET = 72J KET = 36J KET = 48J (2/3 of total) KER = 24J (1/3 of total) KER = 36J Total KE = 72J Total KE = 72J v = 8.5m/s Total KE = 72J v = 6m/s v = 6.9m/s

5. vT = gh 0 0 0 R T g S 0 0 1/2PEg = KET h R T g S 1/2mgh = 1/2mv2 vT Rolls without slipping so vR = vT and KER = KET since it is a ring. Use Energy to fins the velocity of the ring.

6. For the disk 0 0 4gh v = 3 R T g S 0 0 0 R T g S 2/3PEg = KET h 2/3mgh = 1/2mv2 Rolls without slipping so vR = vT and KER = 1/2KET since it is a disk.

7. Systems of objects A Skateboard made of 4 wheels (disks, 1kg each) and a board (2kg). Rolls without slipping at 1m/s So 1/4 in rotation and 3/4 in translation Sort the total energy into buckets: KET = Board and Wheels = 3J KER = Just Wheels KER of disk = 1/2 KET of disk KET of disk = 0.5J so KER disk = 0.25J KER = 1J for all 4 wheels