   Chapter 8, Problem 56P

Chapter
Section
Textbook Problem

A constant torque of 25.0 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 15.0 revolutions. Hint: The angular equivalent of Wnet = FΔx = 1 2 m v f 2 − 1 2 m v i 2 is Wnet = τΔθ = 1 2 I w f 2   −   1 2 I w i 2 . You should convince yourself that this last relationship is correct.

To determine
The angular speed of the grind stone has made 15.0revolutions .

Explanation

Given info: The moment of inertia of the grindstone is 0.130kgm2 , the constant torque of the stone is 25.0Nm , and its angular displacement is 15.0revolutions .

Explanation: From work-energy theorem, the angular speed of the top is calculated that is Wnet=Fs=KEfKEi=12Iωf20 since ωi=0rad/s and the force is F=τΔθ . By using these expressions, the angular speed of the grind stone is determined.

The formula for the angular speed of the angular speed of the grind stone is,

ωf=2τΔθI

• τ is constant torque of the grind stone.
• Δθ is angular displacement.
• I is moment of inertia of the grind stone.

Substitute 25.0Nm for τ , 15.0rev for Δθ , and 0

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