Question

A tire (solid disk) has a mass of 10 kg and a radius of 0.25 m. The tire rests at the top of an incline. When
released, the tire rolls without slipping down to the bottom of the incline. The top of the incline is 10 m in
height above the bottom of the incline.
a) What is the angular velocity of the tire at the bottom of the incline?
b) What would the angular velocity at the bottom of the incline be, if the tire were replaced by a solid
sphere?

Please explain and dont just solve!

Answer 1

This can be solved using conservation of energy Ei PE ; = mgh Ef = KE = 2 mx + 1 1 2 Energy is same migh: mutta į Ico? a) Por closk migh = 1 m 8 Wh +11 mx ² x w² 1 I 22 gh = 0.7582 w² - 7 9.9x10 wash the VoaseVo.a576.2572 w z 45.723 rad/sec. b) gh it is spliese I 2 m² mgh myw027s miraxw? mýth = 6+) marzo gh = 0.2r²w² - wa sh Scanned with 0.77 CamScanner 9.8410 47.32 roads V6.7046-2572