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tensa zangetsu [6.8K]
2 years ago
14

Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if

the system it is rotated about (a) one end of the rod and (b) the center of the rod?
Physics
1 answer:
sergij07 [2.7K]2 years ago
7 0

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

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Answer:

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Explanation:

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Rewriting our equation:

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t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

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3 years ago
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Answer:

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3 years ago
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Answer:

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