A cord is wrapped around the rim of a wheel 0.250m in radius,and a steady pull o

Question

A cord is wrapped around the rim of a wheel 0.250m in radius,and a steady pull of 40.0N is exerted on the cord. The wheel ismounted on frictionless bearings on a horizontal shaft through itscenter. a)The moment of inertia of the wheel about this shaft is5.00 kg. m2 Compute the angular acceleration of thewheel.                         =        rad/s2 A cord is wrapped around the rim of a wheel 0.250m in radius,and a steady pull of 40.0N is exerted on the cord. The wheel ismounted on frictionless bearings on a horizontal shaft through itscenter. a)The moment of inertia of the wheel about this shaft is5.00 kg. m2 Compute the angular acceleration of thewheel.                         =        rad/s2

Explanation / Answer

We know that :           Angular acceleration = / I             Where = Torque                            = Force * perpendicular distance                            = 40.0 N * 0.250 m   = 10 N - m          Moment ofinertia is :                            I = 5.00 kg - m2        =   10 * 5.00 = 50 rad /s2 Hope this helps u! Hope this helps u!

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