Acord is wrapped around the rim of a wheel 0.235 m in radius, and a steady pull of 43.0 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 5.15 kg⋅m2 compute the angular acceleration of the wheel.

Explanation:

given,

radius of wheel = 0.235 m

steady pull = 43 N

moment of inertia of wheel = 5.15 Kg.m²

angular acceleration  = ?

torque = Force x distance

τ = F r

and torque is also known as

τ = I α

computing both the torque expression

I α  = F r

hence, the angular acceleration of wheel is equal to

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