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A uniform hoop rolls without slipping down a 20° inclined plane. What is the acceleration of the hoop's center of mass? The moment of inertia of a uniform solid disk about an axis that passes through its center = mr2 The moment of inertia of a uniform solid disk about an axis that is tangent to its surface = 2mr2. X3.35 m/s2

Respuesta :

Answer:

The acceleration of the hoop's center of mass is 3.35 m/s².

Explanation:

Given that,

Angle = 20°

We need to calculate the acceleration of the hoop's center of mass

Using formula of vertical component

[tex]F = mg\sin\theta[/tex]....(I)

Here, F = ma

Put the value of F in equation (I)

[tex]ma=mg\sin\theta[/tex]

Put the value into the formula

[tex]a=9.8\times\sin20[/tex]

[tex]a=3.35\ m/s^2[/tex]

Hence, The acceleration of the hoop's center of mass is 3.35 m/s².

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