Explanation Given information: The height of the solid cone is 180 mm , the radius of base is 60 mm , rate of spin is 300 rad / s , the rate of precession is 20 rad / s Let us assume the mass of the solid cone is m Write the expression for the height of the center of gravity from the apex. C = 3 4 ( h ) ....... (I) Here, the height of the solid cone is h . Write the expression for the mass moment of inertia. I = 3 10 m r 2 ....... (II) Write the expression for mass moment of inertia in transverse axis. I ′ = 3 20 m r 2 + 3 5 m h 2 = 3 m 5 [ 1 4 r 2 + h 2 ] I ′ m = 3 5 [ 1 4 r 2 + h 2 ] ....... (III) Write the expression for the angular velocity. ω = − ϕ sin θ i + θ j + ( ψ + ϕ cos θ ) k ....... (IV) Here, the angular velocity in x direction is ϕ sin θ the angular velocity in y direction is θ , the angular velocity in is direction is ψ + ϕ cos θ . Write the expression for the angular velocity in x direction. ω x = − ϕ sin θ ....... (V) Write the expression for the angular velocity in y direction. ω y = θ ....... (VI) Write the expression for the angular velocity in z direction. ω z = ψ + ϕ cos θ ....... (VII) Write the expression for the kinetic energy. T = 1 2 ( I x ω x 2 + I y ω y 2 + I z ω z 2 ) ....... (VIII) Let us assume I x = I ′ , I ′ = I y and I z = I . Substitute − ϕ sin θ for ω x , θ ϕ for ω y , ( ψ + ϕ cos θ ) for ω z , I ′ for I x , I ′ for I y and I in Equation (). T = 1 2 ( I ′ ω x 2 + I ′ ω y 2 + I ω z 2 ) = 1 2 I ′ ( ω x 2 + ω y 2 ) + I ω z 2 2 ....... (IX) Divide the Equation () by m . T m = 1 2 I ′ m ( ω x 2 + ω y 2 ) + I ω z 2 2 m ....... (X) Write the expression for the potential energy. V = m g c cos θ V m = g c cos θ ....... (XI) Write the expression for the principle of conservation of energy. E = T + V 2 E m = 2 [ T m + V m ] ....... (XII) Write the expression for the relation between moment of inertia and angular velocity. I ω z = β ω z = β I ....... (XIII) Dividing the Equation (XIII) both side by m . ω z m = β m I β m = I ω z m ....... (XIV) Substitute m g c cos θ for V and 1 2 [ I ′ ( − ϕ sin θ ) 2 + I ′ ( θ ) 2 j + I ω z 2 ] for T in Equation (XII). E = 1 2 [ I ′ ( − ϕ sin θ ) 2 + I ′ ( θ ) 2 j + I ω z 2 ] + m g c cos θ ....... (XV) Write the Expression of momentum. I ′ ϕ sin 2 θ + β cos θ = α ....... (XVI) Divide both side of the Equation (XVI) by m . α m = I ′ m ϕ sin 2 θ + β m cos θ ....... (XVII) Write the expression for the force. F ( x ) = [ 2 E − β 2 i m − 2 g c x ] ( 1 − x 2 ) − m I ′ [ α m − β x m ] 2 = 0 ....... (XVIII) Write the expression for the maximum value of θ . θ max = cos − 1 ( x min ) ....... (XIX) Calculation: Substitute 180 mm for h in Equation (I). C = 3 4 ( 180 mm ) = 3 4 ( 180 mm ) ( 1 m 1000 mm ) = 0.135 m Substitute 60 mm for r in Equation (II). I = 3 10 m ( 60 mm ( 1 m 1000 mm ) ) 2 I m = 3 10 ( 0.06 m ) 2 I m = 1.08 × 10 − 3 m 2 Substitute 60 mm for r and 180 mm for h in Equation (III). I ′ m = 3 5 [ 1 4 ( 60 mm ( 1 m 1000 mm ) ) 2 + ( 180 mm ( 1 m 1000 mm ) ) 2 ] = 3 5 [ 1 4 ( 0 To determine (b) The rate of spin. The rate of precision.

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Author: BEER

Publisher: MCG

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Chapter 18.3, Problem 18.139P

To determine

(a)

The maximum value of θ to ensuring the motion.

To determine

(b)

The rate of spin.

The rate of precision.

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Chapter 18.3, Problem 18.138P

Chapter 18.3, Problem 18.140P

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(a) The maximum value of θ to ensuring the motion.

Solution Summary: The author explains the maximum value of theta to ensuring the motion and the angular velocity.