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A constant torque of 24.5 N · m is applied to a grindstone whose moment of inertia is 0.130 kg · m2. Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 16.9 revolutions. Hint: the angular equivalent of Wnet = FΔx = 1 2 mvf2 − 1 2 mvi2 is Wnet = τΔθ = 1 2 Iωf2 − 1 2 Iωi2. You should convince yourself that this last relationship is correct. (Assume the grindstone starts from rest.)

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Answer:

The final angular speed will be [tex]\omega _f=200rad/sec[/tex]

Explanation:

We have given torque [tex]\tau =24.5N-m[/tex]

Moment of inertia is given [tex]I=0.130kgm^2[/tex]

Angular speed [tex]\Theta =16.9revolutions=16.9\times2\pi =106.132rad[/tex]

Initial angular velocity [tex]\omega _i=0rad/sec[/tex]

Work done is equal to

[tex]w=\frac{1}{2}I\omega _f^2-\frac{1}{2}I\omega _i^2[/tex]

[tex]\tau (\Delta \Theta )=\frac{1}{2}I\omega _f^2[/tex]

[tex]24.5\times 106.132=\frac{1}{2}\times 0.130\times \omega _f^2[/tex]

[tex]\omega _f=200rad/sec[/tex]

So the final angular speed will be [tex]\omega _f=200rad/sec[/tex]

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