Explanation Given information: The weight of the rod ( W ) is 3 lb. The angular velocity ( ω ) is ( 12 rad/s ) i . The angular acceleration ( α ) is − ( 96rad/s 2 ) i . The distance between DF is 9 inch. The distance between AF is 3 inch. Calculation: Show the free body diagram of the system as in Figure (1). The principal axes for the rod CDE are x ′ and y ′ . Calculate angle ( θ ) in triangle C D F using the relation: cos θ = D F C D Substitute 9 in. for DF and 12 in. for CD . cos θ = 9 in . 12 in . θ = cos − 1 ( 0.75 ) θ = 41.41 ° Calculate angular velocity in the x ′ axis ( ω x ′ ) using the relation: ω x ′ = ω cos θ Here, ω is the angular velocity of the system. Substitute 12 rad / s for ω and 41.41 ° for θ . ω x ′ = ( 12 rad / s ) cos ( 41.41 ° ) = 12 × 0.75 = 9 rad / s Calculate angular velocity in the y ′ axis ω y ′ using the relation: ω y ′ = ω sin θ Substitute 12 rad / s for ω and 41.41 ° for θ . ω y ′ = ( 12 rad / s ) sin ( 41.41 ° ) = 12 × 0.66 = 7.9373 rad / s The angular velocity in the z ′ direction is zero. The mass moment of inertia in x ′ direction is zero, because the torque around the principle axis only affects the acceleration. The value of I ¯ x ′ is zero. Write the expression for the mass moment of inertia in the y ′ direction ( I ¯ y ′ ) as follows: I ¯ y ′ = 1 12 m l 2 (1) Here, m is the mass and l is the length of the rod. Calculate the mass of the rod ( m ) using the formula: m = W g Here, g is the acceleration due to gravity. Substitute 3 lb for W and 32.2 ft / s 2 for g . m = 3 lb 32.2 ft / s 2 = 0.09317 lb ⋅ s 2 / ft Find the moment of inertia about y axis I y ′ : Substitute 0.09317 lb ⋅ s 2 / ft for m and 2 ft for l in equation (1). I ¯ y ′ = 1 12 ( 0.09317 lb ⋅ s 2 / ft ) ( 2 ft ) 2 = 0.007764 × 4 = 0.03106 lb ⋅ s 2 ⋅ ft The moment of inertia for I x ′ and I y ′ are same. Calculate the angular momentum at the point D ( H D ) using the formula: H D = I ¯ x ′ ω x ′ i ′ + I ¯ y ′ ω y ′ j ′ + I ¯ z ′ ω z ′ k ′ Substitute 0 for I ¯ x ′ , 9 rad / s for ω x ′ , 0.03106 lb ⋅ s 2 ⋅ ft for I ¯ y ′ , 7.9373 rad / s for ω y ′ , 0 for I ¯ z ′ , and 0 for ω z ′ . H D = ( 0 ) ( 9 rad / s ) i ′ + ( 0.03106 lb ⋅ s 2 ⋅ ft ) ( 7.9373 rad / s ) j ′ + ( 0 ) ( 0 ) k ′ = 0.03106 × 7.9373 = ( 0.24649 lb ⋅ s ⋅ ft ) j ′ Write the expression for the rate of change of angular momentum at point D ( H ˙ D ) as follows: H ˙ D = ( H ˙ D ) D x ′ y ′ z ′ + Ω × H D (2) Here, ( H ˙ D ) D x ′ y ′ z ′ is the rate of change angular momentum with respect to rotating frame D x ′ y ′ z ′ and Ω is the angular velocity of the frame D x ′ y ′ z ′ . Calculate the angular acceleration in the x ′ direction ( α x ′ ) using the relation: α x ′ = α cos θ Here, α is the angular acceleration. Substitute − 96 rad / s 2 for α and 41.41 for θ . α x ′ = ( − 96 rad / s 2 ) cos 41.41 ° = − 96 × 0

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Author: BEER

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Chapter 18.2, Problem 18.61P

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The rate of change (H˙D) of the angular momentum (HD) of the rod.

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Chapter 18.2, Problem 18.60P

Chapter 18.2, Problem 18.62P

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The rate of change ( H ˙ D ) of the angular momentum ( H D ) of the rod.

Solution Summary: The author shows the free body diagram of the system as in Figure (1).

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The rate of change (H˙D) of the angular momentum (HD) of the rod.

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