Explanation Given: The weight of the forked rod, m = 0.4 kg . The velocity of the cylinder, v C = 2 m/s . The radial distance of forked rod, r = ( 0.6 θ ) m . Show the smooth cylinder P using forked rod having a mass m along the vertical slotted path as in Figure (1). Show the direction of the forces as in Figure (2). Determine the angular velocity of the forked rod. v 2 = ( v r ) 2 + ( v θ ) 2 (I). Here, the radial velocity is v r , and the tangential velocity is v θ . Write the expression for the radial acceleration. a r = r ¨ − r θ ˙ 2 (II). Here, radial acceleration is a r and second time derivative of radius ( r ) is r ¨ . Write the expression for the transverse acceleration. a θ = r θ ¨ + 2 r ˙ θ ˙ (III). Here, transverse acceleration is a θ and second time derivative of θ is θ ¨ and first time derivative of radius ( r ) is r ˙ . Determine the angle perpendicular on two components. ψ = tan − 1 ( r d r / d θ ) (IV) Refer Figure (2) Determine the transmission angle φ . φ = 90 ° − ψ (V) Write the formula for equation of motion. ∑ F r = m a r (VI) Here, summation of total force acting on radial direction is ∑ F r mass is m and acceleration radial direction is a r . Refer Figure (2) calculate the components of force. ∑ F r = − N cos φ (VII) Here, the normal force is N Refer Figure (2) write the formula for equation of motion. ∑ F θ = m a θ (VIII). Here, summation of total force acting on transverse direction is ∑ F θ mass is m and acceleration in the transverse direction is a θ . Refer Figure (2) calculate the components of force. ∑ F θ = F − N sin φ (IX). Conclusion: Refer figure (1) The radial distance. r = ( 0.6 θ ) m (X) Differentiate the Equation (X) r ˙ = 0.6 θ ˙ (XI) Again differentiate the Equation (XI) r ¨ = 0.6 θ θ ¨ (XII) Refer figure (2) The radial velocity and tangential velocity components. v r = r ˙ = 0.6 θ ˙ v θ = r θ ˙ = 0.6 θ θ ˙ Substitute 2 m/s 2 for v , 0.6 θ ˙ for v r , and 0.6 θ θ ˙ for v θ in Equation (I) ( 2 ) 2 = ( 0.6 θ ˙ ) 2 + ( 0.6 θ θ ˙ ) 2 2 = 0.36 θ ˙ 2 + 0.36 θ θ ˙ 2 θ ˙ = 2 0.6 ( 1 + θ 2 ) (XIII) Differentiate Equation (I) with respect to time. d d t ( v 2 ) = d d t ( v r ) 2 + d d t ( v θ ) 2 (XIV) Substitute 2 m/s 2 for v , 0.6 θ ˙ for v r , and 0.6 θ θ ˙ for v θ in Equation (XIV) d d t ( 2 2 ) = d d t ( 0.6 θ ˙ ) 2 + d d t ( 0.6 θ θ ˙ ) 2 0 = [ 2 × 0.36 θ ˙ × θ ¨ + ( 2 × 0.36 × θ θ ˙ 3 + 2 × 0.36 × θ 2 θ ˙ θ ¨ ) ] 0 = 0.72 θ ˙ θ ¨ + 0.72 θ ˙ ( θ θ ˙ 2 + θ 2 θ ¨ ) (XV) Now divide by 1 0.72 θ ˙ in Equation (XV) 0 = θ ¨ + θ θ ˙ 2 + θ 2 θ ¨ θ θ ˙ 2 = − ( 1 + θ 2 ) θ ¨ θ ¨ = − θ θ ˙ 2 1 + θ 2 (XVI) Refer figure (2)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.6, Problem 102P

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The normal force N of the rod on the particle.

The force F of the rod on the particle.

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Chapter 13.6, Problem 101P

Chapter 13.6, Problem 103P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The normal force N of the rod on the particle. The force F of the rod on the particle.

Solution Summary: The author explains how to determine the normal force and the force of the rod on the particle.