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A ship's wheel has a moment of inertia of 0.994 kilogram·meters squared. The inner radius of the ring is 25 centimeters, and the outer radius of the ring is 35 centimeters. Disagreeing over which way to go, the captain and the helmsman try to turn the wheel in opposite directions. The captain applies a force of 310 newtons at the inner radius, while the helmsman applies a force of 286 newtons at the outer radius. What is the magnitude of the angular acceleration of the wheel? Include units in your answer. Answer must be in 3 significant digits.

User Micycle
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1 Answer

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Given,

The moment of inertia of the wheel, I=0.994 kg·m²

The inner radius of the ring, r=25 cm=0.25 m

The outer radius of the ring, R=35 cm=0.35 m

The force applied by the captain, F=310 N

The force applied by the helmsman, f=286 N

The torque is the measure of the rotation caused by an applied force.

The torque of the wheel is given by,


\tau=fR-Fr=I\alpha

Where α is the angular acceleration of the wheel.

On substituting the known values,


\begin{gathered} 286*0.35-310*0.25=0.994*\alpha \\ \Rightarrow\alpha=(286*0.35-310*0.25)/(0.994) \\ =22.7\text{ rad/s}^2 \end{gathered}

Therefore the angular acceleration of the wheel is 22.7 rad/s²

User Sayan Pathak
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