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A cord is wrapped around the rim of a wheel 0.240 m in radius, and a steady pull of 41.5 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 4.85 kg⋅m2.Compute the angular acceleration of the wheel.

Respuesta :

Given

Radius is r=0.240 m

The force is F=41.5 N

Moment of inertia,

[tex]I=4.85\text{ kg.m}^2[/tex]

To find

The angular acceleration

Explanation

We have

[tex]\begin{gathered} F.r=I\alpha \\ \Rightarrow41.5\times0.240=4.85\alpha \\ \Rightarrow\alpha=\frac{2.05rad}{s^2} \end{gathered}[/tex]

Conclusion

The angular acceleration is

[tex]2.05\text{ rad/s}^2[/tex]

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