(a) Explanation Given information: The mass of the wheel ( m ) is 18 kg. The velocity of the wheel ( N ) is 12.5 rev/s. The force exerted in the system ( F ) is ( 160 N ) j . The couple exerted in the system ( M C ) is ( 14.7 N ⋅ m ) k . The height from center of machine to B ( y B ) is 182 mm. The height from center of machine to D ( y D ) is -182 mm. The horizontal distance from center of machine to B ( x B ) is 75 mm. The horizontal distance from center of machine to D ( x D ) is -75 mm. Calculation: Calculate the angular velocity ( ω ) using the formula: ω = 2 π N Substitute 12.5 rev/s for N . ω = 2 π ( 12.5 ) = ( 78.54 rad/s ) i Calculate the force system exerted by the wheel on the machine ( F ) using the relation: F = − F ′ Substitute ( 160 N ) j for F . F ′ = − ( 160 N ) j Calculate the force system exerted on the wheel ( F ′ ) using the formula: F ′ = − m r ¯ ω 2 Substitute 18 kg for m , − ( 160 N ) j for F ′ and 78.54 rad/s for ω . − 160 = − 18 × r ¯ ( 78.54 ) 2 r ¯ = 160 18 ( 78.54 ) 2 r ¯ = 1.441 × 10 − 3 m × 1 , 000 mm m r ¯ = 1.441 mm Write the expression for the angular momentum about the center of mass as follows: H C = I x ω i − I x y ω j − I x z ω k Here, H C is the angular momentum about the center of mass, I ¯ x is the moment of inertia in x plane, I ¯ x y is the moment of inertia in xy plane, I ¯ x z is the moment of inertia in xz plane and ω is the angular velocity. Calculate the couple system exerted by the wheel on the machine ( M C ) using the relation: M C = − M ′ C Substitute ( 14.7 N ⋅ m ) k for M C (b) To determine The mass at the point A ( m A ) , B ( m B ) , D ( m D ) and E ( m E ) .

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ISBN: 9781260663778

Author: BEER

Publisher: MCG

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Chapter 18.2, Problem 18.70P

(a)

To determine

The distance from the axis of rotation (r¯) to the mass center of the wheel and the products of inertia (Ixy) and (Izx).

(b)

To determine

The mass at the point A (mA), B (mB), D (mD) and E (mE).

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Chapter 18.2, Problem 18.69P

Chapter 18.2, Problem 18.71P

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Chapter 18 Solutions

VECTOR MECH. FOR EGR: STATS & DYNAM (LL

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 distance from the axis of rotation ( r ¯ ) to the mass center of the wheel and the products of inertia ( I x y ) and ( I z x ) .

Solution Summary: The author calculates the distance from the axis of rotation to the mass center of the wheel and the products of inertia.

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The distance from the axis of rotation (r¯) to the mass center of the wheel and the products of inertia (Ixy) and (Izx).

The mass at the point A (mA), B (mB), D (mD) and E (mE).

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Chapter 18 Solutions