This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: From part b), velocity of the block is v = 3.06 m/s after it falls a distance d = 2.00 m. Applying one of the kinematics equations, we have v 2 – 0 2 = 2ad ⇒ a = v 2 /(2d) = 2.34 m/s 2 b) What is the speed of the object when it has fallen a distance 2.00 m after being released from rest? Only the gravitational force exerted on the block does work on the system. Hence the mechanical energy of the system is conserved. The energy conservation equation yields mgd = (1/2)mv 2 + (1/2)I S ω S 2 + (1/2)I P ω P 2 where d = 2 m. Since the cord does not slip on the sphere or pulley, the angular velocity of the sphere and pulley are related to the translational velocity of the block as v = R ω S = r ω P Expressing ω S and ω P in terms of v and plugging them into the energy conservation equation, the speed of the block after the block falls distance d is mgd = (1/2)mv 2 + (1/2)(2/5)MR 2 (v/R) 2 + (1/2)I P (v/r) 2 v = √ [2mgd/((2/5)M+ I P /r 2 + m )] = 3.06 m/s...
View Full Document
This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08