Explanation Given information: The angular velocity of the disk is ω i and the angular acceleration is α i . The below figure represent the schematic diagram of the disk. Figure-(1) Write the expression of moment of inertia about x ′ axis. I x ′ = 1 4 m r 2 Here, the mass is m and the radius is r . Write the expression of moment of inertia about y ′ axis. I y ′ = 1 2 m r 2 Write the expression of moment of inertia about z ′ axis. I z ′ = 1 4 m r 2 Write the expression of angular acceleration about x ′ axis. ω x ′ = − ω sin β Here, the mass angular acceleration is ω and the angle is β . Write the expression of angular acceleration about y ′ axis. ω y ′ = ω cos β Write the expression of angular acceleration about z ′ axis. ω z ′ = 0 Write the expression of angular momentum. H G = I x ′ ω x ′ i ′ + I y ′ ω y ′ j ′ + I z ′ ω z ′ k ′ ........ (I) Here, the unit vector along the x ′ axis is i ′ , the unit vector along the y ′ axis is j ′ and the unit vector along the z ′ axis is k ′ . Write the expression of unit vector along x ′ axis. i ′ = i cos β − j sin β Write the expression of unit vector along y ′ axis. j ′ = i sin β − j cos β Substitute i cos β − j sin β for i ′ , i sin β − j cos β for j ′ , ω cos β for ω y ′ , − ω sin β for ω x ′ , 1 4 m r 2 for I z ′ , 1 2 m r 2 for I y ′ , 1 4 m r 2 for I x ′ and 0 for ω z in Equation (I). H G = [ ( 1 4 m r 2 ) ( − ω sin β ) ( i cos β − j sin β ) + ( 1 2 m r 2 ) ( ω cos β ) ( i sin β − j cos β ) + I z ′ ( 0 ) k ′ ] = ( 1 4 m r 2 ω ) [ ( − sin β ) ( i cos β − j sin β ) + 2 ( cos β ) ( i sin β − j cos β ) ] = ( 1 4 m r 2 ω ) [ ( − sin β cos β i + sin 2 β j ) + ( 2 sin β cos β i − j 2 cos 2 β ) ] = ( 1 4 m r 2 ω ) [ sin β cos β i + ( 1 + cos 2 β ) j ] Write the expression of momentum about G in G x y z frame. ( H G ) G x y z = 1 4 m r 2 α [ sin β cos β i + ( 1 + cos 2 β ) j ] Here, the angular acceleration is α . Write the expression of rate of change of H ˙ G of the angular momentum H G of the disk

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Chapter 18.2, Problem 18.64P

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The rate of change of H˙G of the angular momentum HG of the disk.

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Chapter 18.2, Problem 18.63P

Chapter 18.2, Problem 18.65P

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Ch. 18.1 - Prob. 18.1P 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 the...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Solve Prob. 18.6, assuming that the solid...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P

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The rate of change of H ˙ G of the angular momentum H G of the disk.

Solution Summary: The author explains the rate of change of stackreldotH_G of the angular momentum and the moment of inertia.

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The rate of change of H˙G of the angular momentum HG of the disk.

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