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A mass m on a frictionless plane, inclined at an angle θ with respect to the horizontal, is connected by a cord that runs parallel to the inclined plane and is wrapped around a flywheel of radius r and moment of inertia

[tex]i=\frac{3mr^{2}}{4}[/tex]

what is the acceleration of the mass down
the plane?

Sagot :

onyebuchinnaji

The acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

Conservation of angular momentum

The acceleration of the mass down the plane is determined by applying the principle of conservation of angular momentum.

Fr = Iα

where;

  • F is weight of the object parallel to the plane
  • r is the radius of the flywheel
  • I is moment of inertia
  • α is angular acceleration

(mg sinθ)r = Iα

(mg sinθ)r = I(ar)

(mg sinθ) = I(a)

[tex]a = \frac{mg \times sin(\theta)}{I} \\\\a = \frac{mg \times sin(\theta)}{3mr^2/4} \\\\a = \frac{4mg \times sin(\theta)}{3mr^2}[/tex]

Thus, the acceleration of the mass down the plane is determined as (4mg sinθ)/(3mr²).

Learn more about acceleration here: https://brainly.com/question/14344386

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