(a) To determine The expression for the speed v ′ of the cylinder. The percentage n of the cylinder kinetic energy lost if θ = 10 ° and v 0 = 0.25 v is 0.96 % . Explanation The schematic diagram of the system is shown below. Figure-(1) Write the expression for the initial angular velocity. ω = v r (I) Here, the initial angular velocity of the cylinder is ω , the velocity of the cylinder before the impact is v and the radius of the cylinder is r . Write the expression for the final angular velocity. ω ′ = v ′ r (II) Here, the final angular velocity of the cylinder is ω ′ and the velocity of the cylinder after the impact is v ′ . Write the expression for the mass moment of inertia of the cylinder about point G . I G = 1 2 m r 2 (III) Here, the mass moment of inertia of the cylinder about point G is I G and the mass of the cylinder is m . Write the expression for the mass moment of inertia of the cylinder about point B . I B = 1 2 m r 2 + m r 2 = 3 2 m r 2 (IV) Here, the mass moment of inertia of the cylinder about point B is I B . Write the expression for the initial angular moment of the cylinder about point B . ( H B ) 1 = I G ω + m ( v − v 0 ) r cos θ (V) Here, the initial angular moment of the cylinder about point B is ( H B ) 1 , the travelling speed of the ramp is v 0 and θ is the angle of inclination of the ramp. Write the expression for the final angular moment of the cylinder about point B . ( H B ) 2 = I B ω ′ (VI) Here, the final angular moment of the cylinder about point B is ( H B ) 2 . Write the expression for the conservation of angular momentum. ( H B ) 1 = ( H B ) 2 (VII) Write the expression for the velocity of cylinder about point G using the cosine law. v G 2 = v 0 2 + ( v ′ ) 2 − 2 v 0 v ′ cos ( 180 ° − θ ) (VIII) Here, the velocity of cylinder about point G is v G . Write the expression for the mass moment of inertia of the cylinder about point A . I A = 1 2 m r 2 + m r 2 = 3 2 m r 2 (IX) Here, the mass moment of inertia of the cylinder about point A is I A . Write the expression for the initial kinetic energy of the cylinder. T 1 = 1 2 I A ω 2 (X) Here, the initial kinetic energy of the cylinder is T 1 . Write the expression for the final kinetic energy of the concrete block. T 2 = 1 2 I G ( ω ′ ) 2 + 1 2 m v G 2 (XI) Here, for the final kinetic energy of the concrete block is T 2 . Write the expression for the percentage loss n of energy due to the impact. n = T 1 − T 2 T 1 × 100 % (XII) Here, the percentage loss of energy due to the impact is n . Conclusion: Substitute ( I G ω + m ( v − v 0 ) r cos θ ) for ( H B ) 1 and I B ω ′ for ( H B ) 2 in Equation (VII). I G ω + m ( v − v 0 ) r cos θ = I B ω ′ (XIII) Substitute 1 2 m r 2 for I G , v / r for ω , 3 2 m r 2 for I B and v ′ / r for ω ′ in Equation (XIII) (b) To determine The percentage n of the cylinder kinetic energy lost if θ = 10 ° and v 0 = 0.5 v is 17.28 % .

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Chapter 6.9, Problem 202RP

(a)

To determine

The expression for the speed v′ of the cylinder.

The percentage n of the cylinder kinetic energy lost if θ=10° and v0=0.25v is 0.96%.

(b)

To determine

The percentage n of the cylinder kinetic energy lost if θ=10° and v0=0.5v is 17.28%.

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Chapter 6.9, Problem 201RP

Chapter 6.9, Problem 203RP

Chapter 6 Solutions

Engineering Mechanics: Dynamics

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The expression for the speed v ′ of the cylinder. The percentage n of the cylinder kinetic energy lost if θ = 10 ° and v 0 = 0.25 v is 0.96 % .

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The expression for the speed v′ of the cylinder. The percentage n of the cylinder kinetic energy lost if θ=10° and v0=0.25v is 0.96%.

The percentage n of the cylinder kinetic energy lost if θ=10° and v0=0.5v is 17.28%.

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Chapter 6 Solutions

The expression for the speed v′ of the cylinder.

The percentage n of the cylinder kinetic energy lost if θ=10° and v0=0.25v is 0.96%.