Explanation Given information: The mass ( m ) of the top is 85 g. The radius of gyration ( k x 2 ) along x -axis is 45 mm. The radius of gyration ( k z 2 ) along y -axis is 21 mm. The distance of mass center ( c ) from O is 37.5 mm. The rate of spin ( ψ ˙ ) of the top about its axis of symmetry is 1,800 rpm. Calculation: Consider principal axes x , y , z with origin at O . Sketch the free body diagram of the system with principal axes as shown in Figure (1). The rectangular component of the rate of precession acting in the x -axis and z -axis is φ ˙ sin θ and φ ˙ cos θ . Write the equation of angular velocity of the system ( ω ) . ω = ω x i + ω y j + ω z k Here, ω x angular velocity in the x -axis, ω y is the angular velocity in the y -axis, and ω z is the angular velocity in the z -axis. Substitute φ ˙ sin θ for ω x , 0 for ω y , and φ ˙ cos θ + ψ ˙ for ω z . ω = φ ˙ sin θ i + ( ψ ˙ + φ ˙ cos θ ) k (1) Here, φ ˙ is the rate of precession, θ is the φ ˙ , angle between rate of precession and rate of spin, and ψ ˙ is the rate of spin. Write the magnitude of x -component of angular velocity ( ω x ) of the system from equation (1) ω x = φ ˙ sin θ Write the magnitude of x -component of angular velocity ( ω z ) of the system from equation (1) ω z = ( ψ ˙ + φ ˙ cos θ ) k The moment of inertia ( I y ) in the y -axis is zero. Write the angular momentum about O ( H O ) . H O = I ′ ω x i + I y ω y j + I ω z k Here, I ′ is the moment of inertia in x -axis, I y is the moment of inertia in the y -axis, and I is the moment of inertia in the z -axis. Substitute 0 for I y . H O = I ′ ω x i + ( 0 ) ω y j + I ω z k = I ′ ω x i + I ω z k Write the equation of the angular velocity of the reference frame as Oxyz . Ω = φ ˙ sin θ i + φ ˙ cos θ k The rate of change of angular momentum ( H ˙ O ) O x y z about the reference frame is 0. Write the equation of the rate of change of angular momentum about O . H ˙ O = ( H ˙ O ) x y z + Ω × H O Substitute 0 for ( H ˙ O ) x y z , φ ˙ sin θ i + φ ˙ cos θ k for Ω and I ′ ω x i + I ω z k for H ˙ O . H ˙ O = ( 0 ) + ( φ ˙ sin θ i + φ ˙ cos θ k ) × ( I ′ φ ˙ sin θ i + I ω z k ) = 0 + I ′ φ ˙ 2 sin θ cos θ j − I ω z φ ˙ sin θ j + 0 = ( I ′ φ ˙ sin θ cos θ − I ω z sin θ ) φ ˙ j (2) Refer Figure (1), Take moment about O ( M O ) . M O = − W c sin θ j (3) Here, W is the weight and c sin θ is the horizontal distance between O and G . Prove the condition for steady precession. The moment about O is equal to rate of change of angular momentum about O . Equate Equations (1) and (2). − W c sin θ j = ( I ′ φ ˙ sin θ cos θ − I ω z sin θ ) φ ˙ j − W c = ( I ′ φ ˙ cos θ − I ω z ) φ ˙ W c = ( I ω z − I ′ φ ˙ cos θ ) φ ˙ (4) Substitute ψ ˙ + φ ˙ cos θ for ω z

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ISBN: 9781260804638

Author: BEER

Publisher: MCG

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Moment of Inertia

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Chapter 18.3, Problem 18.109P

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The two possible rates of steady precession corresponding to θ=30°.

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Chapter 18.3, Problem 18.108P

Chapter 18.3, Problem 18.110P

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Chapter 18 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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The two possible rates of steady precession corresponding to θ = 30 ° .

Solution Summary: The author illustrates the two possible rates of steady precession corresponding to theta =30°.

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