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14 October, 23:04

A hollow, thin-walled sphere of mass 14.0 kg and diameter 49.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by θ (t) = At2+Bt4, where A has numerical value 1.10 and B has numerical value 1.60.

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  1. 15 October, 01:21
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    Answer

    Given,

    θ (t) = At² + Bt⁴

    where A = 1.10

    B = 1.60

    a) calculating the unit of A and B

    Unit of A = unit of θ / unit of t²

    = rad/s²

    Unit of B = unit of θ / unit of t⁴

    = rad/s⁴

    (b)

    calculating angular speed at time t = 3 s

    mass m = 14 kg

    diameter d = 49 cm

    radius r = d / 2 = 24.5 cm = 0.245 m

    Angular momentum of the sphere

    L = I w

    I = moment of inertia

    I = (2 / 3) m r²

    I = 0.667 x 14 x 0.245²

    I = 0.5605 kg m²

    angular speed

    ω = dθ / dt

    = 2At + 4Bt^3

    = (2 x 1.10 x 3) + (4 x 1.6 x 3³)

    = 179.4 rad / s

    So, angular momentum

    L = 0.5605 x 179.4

    L = 100.55 kg m²/s

    (c) now, calculating net torque

    T = I α

    α = angular acceleration = dω / dt

    = 2 A + 12 B t²

    = 2 x 1.1 + 12 x 1.6 x 3²

    = 175 rad / s²

    T = 175 x 0.5605

    T = 98.08 N m
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