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 Nm
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)

Physics  John, Sunday, March 13, 2016 at 4:00pm
2.00

Physics  Anonymous, Tuesday, February 14, 2017 at 2:15am
n
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