6. The pulley in the system at right is a uniform disk (I = ½ MR\") with a mass

Question

6. The pulley in the system at right is a uniform disk (I = ½ MR") with a mass of 2m. 3m The block of mass 3m is on a frictionless table top. Using conservation of mechanical energy, what is the speed of the blocks when the hanging mass has fallen a distance h? a. b. The gravitational potential energy lost by the hanging mass is divided between the kinetic energy of the blocks and the pulley. What fraction of the KE goes to each? c. If the pulley were replaced with ring, would it get a larger or smaller fraction of the KE?

Explanation / Answer

(A) Applying energy conservation,

Pei + KEi = PEf + KEf

m g h + 0 = 0 + ((3m + m)v^2/2)+((2M)R^2 / 2)(v/R)^2 /2

M gh = 2 M v^2 + M v^2 / 2

g h = (5/2) v^2

v = sqrt(2 g h / 5)

(B) total KE = 2.5 M v^2

  
FOr 3M block: KE = 1.5 M v^2

fraction = 1.5/2.5 = 0.6

for M block: KE = 0.5 M v^2

fraction = 0.5/2.5 = 0.2

For pulley: KE = 0.5 M v^2

fraction = 0.2

(C) for ring I is greater.

so fraction of KE will be larger now.

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.