s=2.08 m
m=0.592 kg
R=0.0786 m
a is the acceleration of the system (m1+m2)
m₁a= T₁
m₂a= m₂g-T₂
Iε=M => mR²•a/2R = (T₂-T₁)R,
T₂-T₁= ma/2,
a= m₂g/( m₁+m₂+m/2)=...
a=v²/2s
v=sqrt(2as)=...
ω=v/R=...
The system shown in the figure below consists of a m1 = 5.52-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m2 = 2.76-kg block.The pulley is a uniform disk of radius 7.86 cm and mass 0.592 kg. Calculate the speed of the m2 = 2.76-kg block after it is released from rest and falls a distance of 2.08 m. Calculate the angular speed of the pulley at this instant.
1 answer
1 answer