Consider the system shown in the figure below with m1 = 27.0 kg, m2 = 13.4 kg, R

Question

Consider the system shown in the figure below with m1 = 27.0 kg, m2 = 13.4 kg, R = 0.110 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 5.00 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley. (a) Calculate the time interval required for m1 to hit the floor. Delta t 1 = 1 S (b) How would your answer change if the pulley were massless? Delta t 2 = 2 S

Explanation / Answer

PE1 = KE1 + KE2 + KEp + PE2

PE1 = m1 g h , PE2 = m2 g h

KE1 = (1/2) m1 * v2 , KE2 = (1/2) m2 * v2

KEp = (1/2) I w2 , I = (1/2) M r2

PE1 = KE1 + KE2 + KEp + PE2

m1 g h = (1/2) m1 * v2 + (1/2) m2 * v2 + (1/2) I w2 + m2 g h

substitute in I = (1/2) M r2 , w = v/r

m1 g h = (1/2) m1 * v2 + (1/2) m2 * v2 + (1/2) (1/2) M r2 (v/r)2 + m2 g h

m1 g h = (1/2) m1 * v2 + (1/2) m2 * v2 + (1/4) M v2 + m2 g h

m1 g h - m2 g h = v2[(1/2) m1+ (1/2) m2+ (1/4) M]

v2 = (m1 g h - m2 g h) / [(1/2) m1 + (1/2) m2 + (1/4) M]

v2= g * h * (m1 - m2) / [(1/2) m1 + (1/2) m2 + (1/4) M]

v = sqrt [g * h * (m1 - m2) / [ (1/2) m1 + (1/2) m2 + (1/4) M ]

v = sqrt [9.81 * 5.00 * (27 kg - 13.4 kg) / [(1/2) 27 kg + (1/2) 13.4 kg + (1/4) 5.00 kg] ]

v = 5.57 m/s

h = h0 + (1/2) * (v - v0) * t

h = v * t / 2

t = 2 * h / v = 2 * 5.00 m / 5.57 m/s

t = 1.795 sec

---------------------------------------------------

v = sqrt [ g * h * (m1 - m2) / [(1/2) m1 + (1/2) m2]]

v = sqrt [9.81 * 5.00 * (27 kg - 13.4 kg) / [(1/2) 27 + (1/2) 13.4 kg]]

v = 5.74 m/s

t = 2 * h / v = 2 * 5.00 m / 5.74 m/s

t = 1.74 sec

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---