A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and o

Question

A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. What is the wheels angular acceleration? What force does the axle apply to the wheel? What torque does the axle apply to the wheel? A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. What is the wheels angular acceleration? What force does the axle apply to the wheel? What torque does the axle apply to the wheel?

Explanation / Answer

We Know

Torque = Moment of Inertia * Angular Acceralation ...1

Also Torque = Force * Radius . ..2

a. Moment of Inertia of Solid Disc = .5 m * r *r ...3

Equating 1 and 2 and putting values

,5 m x r x r x A = F x r

=> A = (32/(0.5 x8.8 x 2.8))

Angular accelaration A = 2.5974025974 radian/s^-2

B. Since no linear Accelaration will be there, using Newton's third law, Force by Axle = 32N

C. Since the force by Axle is imparted at center of Axle, Torque = 32 x 0 = 0

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