Concept explainers
(a)
Velocity of cylinder A.
Answer to Problem 17.14P
Velocity of cylinder A is,
Explanation of Solution
Given information:
Mass of block (mb) = 15kg.
Mass of cylinder (ma) = 5kg.
Radius of gyration (k) = 160mm.
Coefficient of friction between block and surface (
External force 0n the cylinder (P) = 200N.
Radius of outer pulley (ra) = 250mm.
Radius of inner pulley (rb) = 150mm.
Distance moved by the cylinder A (sa) = 1m.
Calculation:
Let radius of inner pulley is rb and radius of outer pulley is ra.
Velocity of block B attached to the inner pulley is vb and velocity of cylinder A is va.
Velocity ratio
Distance moved by the cylinder A is sa and distance moved by the block B is sb.
So, displacement ratio
For initial condition; Angular velocity is zero. So, initial kinetic energy
E1=0
Final kinetic energy is the summation of kinetic energy of cylinder A, kinetic energy of block B and kinetic energy of double pulley.
Equlibrium force in normal direction.
Frictional force
Work done by the system
Substitute the value of E1, E2 and W in work energy equation.
(b)
Total distance moved by block B.
Answer to Problem 17.14P
Total distance moved by block B is s = 1.936 m.
Explanation of Solution
Given information:
Mass of block (mb) = 15kg.
Mass of cylinder (ma) = 5kg.
Radius of gyration (k) = 160mm.
Coefficient of friction between block and surface (
External force 0n the cylinder (P) = 200N.
Radius of outer pulley (ra) = 250mm.
Radius of inner pulley (rb) = 150mm.
Distance moved by the cylinder A (sa) = 1m.
Calculation:
Velocity of block B
Velocity of block B attached to the inner pulley is vb and velocity of cylinder A is va.
Angular velocity of the pulley
Kinetic energy after cylinder A strike the ground is the summation of kinetic energy of block B and kinetic energy of double pulley.
Kinetic energy after the system comes to rest is zero.
E4=0
Work done by the system
Substitute the value of E1, E2 and W in work energy equation
Total distance moved by the block B is
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Chapter 17 Solutions
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