Answered

IDNLearn.com: Your trusted platform for finding reliable answers. Our experts are ready to provide prompt and detailed answers to any questions you may have.

A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the outer radius of the ring is 32 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 314 newtons at the inner radius, while the helmsman applies a force of 290 newtons at the outer radius. What is the magnitude of the angular acceleration of the wheel?

Sagot :

We can use the formula of the moment of inertia given by:

[tex]r\cdot F=I\alpha[/tex]

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

[tex]\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}[/tex]

Answer:

12 rad/s²

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.