(a) Explanation Given information: The height of a solid cone ( h ) is 180 mm. The radius of a circular base ( r ) is 60 mm. The position of the cone ( θ 0 ) is 30 ° . The spin rate ( ψ ˙ 0 ) is 300 rad/s . The rate of precession ( ϕ ˙ 0 ) is 20 rad/s . The rate of nutation is zero. Calculation: Determine the value of ( I m ) using the relation. I m = 3 10 r 2 Substitute 60 mm for r . I m = 3 10 ( 60 mm × 1 m 1 , 000 mm ) 2 = 1.08 × 10 − 3 m 2 Determine the value of ( I ′ m ) using the relation. I ′ m = 3 5 ( 1 4 r 2 + h 2 ) Substitute 60 mm for r and 180 mm for h . I ′ m = 3 5 ( 1 4 × ( 60 mm × 1 m 1 , 000 mm ) 2 + ( 180 mm × 1 m 1 , 000 mm ) 2 ) = 19.98 × 10 − 3 m 2 Determine the initial angular velocity along x –axis ω x using the relation. ω x = − ϕ ˙ 0 sin θ 0 Substitute 20 rad/s for ϕ ˙ 0 and 30 ° for θ 0 . ω x = − 20 sin 30 ° = − 10 rad/s Determine the initial angular velocity along y –axis ω y using the relation. ω y = θ ˙ 0 Substitute 0 for θ ˙ 0 . ω y = 0 Determine the initial angular velocity along z –axis ω z using the relation. ω z = ψ ˙ 0 + ϕ ˙ 0 cos θ 0 Substitute 300 rad/s for ψ ˙ 0 , 20 rad/s for ϕ ˙ 0 , and 30 ° for θ 0 . ω z = 300 + 20 cos 30 ° = 317.32 rad/s Determine the value of ( T m ) using the relation. ( T m ) = 1 2 I ′ m ( ω x 2 + ω y 2 ) + 1 2 I m ( ω z 2 ) Here, T is the kinetic energy. Substitute 19.98 × 10 − 3 m 2 for I ′ m , − 10 rad/s for ω x , 0 for ω y , 1.08 × 10 − 3 m 2 for I m , and 317.32 rad/s for ω z . ( T m ) = 1 2 ( 19.98 × 10 − 3 ) ( ( − 10 ) 2 + 0 ) + 1 2 × ( 1.08 × 10 − 3 ) ( 317.32 ) 2 = 55.372 m 2 /s 2 Determine the value of distance c using the relation. c = 3 4 h Substitute 180 mm for h . c = 3 4 × ( 180 mm × 1 m 1 , 000 mm ) = 0.135 m Determine the value of ( V m ) using the relation. V m = g c cos θ 0 Here, V is the potential energy and g is the acceleration due to gravity. Substitute 9.81 m/s 2 for g , 0.135 m for c , and 30 ° for θ 0 . V m = 9.81 × 0.135 cos 30 ° = 1.147 m 2 /s 2 Determine the value of ( 2 E m ) using the relation. 2 E m = 2 ( T m + V m ) Here, E is a constant. Substitute 55.372 m 2 /s 2 for T m and 1 (b) To determine The precession rate and the spin rate values.

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

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Forced undamped vibrations

In mechanical engineering, vibration is expressed as the term that occurs due to an object's backward and forward movement about a point of equilibrium. Generally, there are three types of vibrations which are,

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Chapter 18.3, Problem 18.139P

(a)

To determine

The maximum value of θ in the ensuing motion.

(b)

To determine

The precession rate and the spin rate values.

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Chapter 18.3, Problem 18.138P

Chapter 18.3, Problem 18.140P

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Chapter 18 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

Ch. 18.1 - A thin, homogeneous disk of mass m and radius r...Ch. 18.1 - Prob. 18.2P Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - A homogeneous disk of mass m = 8 kg rotates at the...Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P Ch. 18.1 - Determine the angular momentum HO of the disk of...

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The maximum value of θ in the ensuing motion.

Solution Summary: The author calculates the value of (Im) using the relation.

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The maximum value of θ in the ensuing motion.

The precession rate and the spin rate values.

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Chapter 18 Solutions