The magnitude and direction of the angular momentum about the center of the disk is 0.996 kgm²/s.
The angular acceleration of the disk is determined by applying the principle of conservation of angular momentum as shown below;
Fr = Iα
α = Fr/I
Where;
I = ¹/₂MR²
I = ¹/₂ x (0.5) x (0.08)²
I = 1.6 x 10⁻³ kgm²
[tex]\alpha = \frac{14 \times 0.22}{1.6\times 10^{-3}}\\\\\alpha = 1,925 \ rad/s^2[/tex]
ωf = ωi + αt
ωf = 45 + (1925 x 0.3)
ωf = 622.5 rad/s
L = Iω
L = (1.6 x 10⁻³) x (622.5)
L = 0.996 kgm²/s
Thus, the magnitude and direction of the angular momentum about the center of the disk is 0.996 kgm²/s.
The complete question is below:
A uniform solid disk with radius 8 cm has mass 0.5 kg (moment of inertia I = (1/2)MR²). A constant force 14 N is applied as shown. At the instant shown, the amount of string that has wound off the disk is d = 22 cm, and the angular velocity of the disk is 45 radians/s into the page.
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