Physics

A cord is wrapped around the rim of a wheel .250 m in radius, and a steady pull of 40.0 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 5.00kg*m^2. Compute the angular acceleration of the wheel.

The torque is T = 40 N x 0.25 M = 10 N-m
The moment of inertia is I = 5.00 kg/m^2
The angular acceleration in radians/s^2
alpha = T/I
Do the numbers

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asked by Karla
  1. R1R2F/(I+mR2square)

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    posted by hi
  2. 2.00

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    posted by John

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