Posted by **Karla** on Tuesday, March 27, 2007 at 12:00pm.

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

- Physics -
**hi**, Thursday, December 6, 2012 at 9:02pm
R1R2F/(I+mR2square)

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