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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

User Pretseli
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1 Answer

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Answer:

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

Step-by-step explanation:

Given that,

Angle = 20°

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

Using formula of vertical component


F = mg\sin\theta....(I)

Here, F = ma

Put the value of F in equation (I)


ma=mg\sin\theta

Put the value into the formula


a=9.8*\sin20


a=3.35\ m/s^2

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

User Eugene Nacu
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