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A force FT is applied to a cord wrapped around a pulley with moment of inertia I = 0.435 kg·m2 and radius R0 = 33.0 cm. (See the figure.) There is a frictional torque τfr at the axle of 1.10 m·N. Suppose the force FT is given by the relation FT = 4.00t - 0.10t2 (newtons) where t is in seconds. If the pulley starts from rest, what is the linear speed of a point on its rim 9.0 s later?

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  1. a = angular acceleration
    I = rotational inertia
    w = angular velocity
    v = linear velocity
    Torque = RxFT = Ia
    so,
    a = RxFT/I = (R/I)(4t-0.10t^2)
    w = integral(a) with respect to t
    v= wR

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