Answer:
The moment of inertia I is
I = 2.205x10^-4 kg/m^2
Explanation:
Given mass m = 0.5 kg
And side lenght = 0.03 m
Moment of inertia I = mass x radius of rotation squared
I = mr^2
In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.
Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m
Therefore,
I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2
(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
<h3> Linear acceleration of the yoyo</h3>The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
rT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
Torque equation for frictional force;
solve (1) and (2)
since the yoyo is pulled in vertical direction, T = mg
α = a/R
α = 3.21/0.04
α = 80.25 rad/s²
<h3>Weight of the yoyo</h3>W = mg
W = 0.15 x 9.8 = 1.47 N
<h3>Tension in the rope </h3>T = mg = 1.47 N
<h3>Angular speed of the yoyo </h3>v² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
Learn more about angular speed here: brainly.com/question/6860269
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