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Triss [41]
3 years ago
6

A large stick is pivoted about one end and allowed to swing back and forth with no friction as a physical pendulum. The mass of

the stick is 4.8 kg and its center of gravity (found by finding its balance point) is 1.4 m from the pivot. If the period of the swinging stick is 9 seconds, what is the moment of inertia of the stick about an axis through the pivot
Physics
1 answer:
Tamiku [17]3 years ago
5 0

Answer:

I = 94.33 kg m^2

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

\tau = mg dsin\theta

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

Inertia = I

now we have

I \alpha = mg d sin\theta

now for small angular displacement we will have

\alpha = \frac{mgd}{I}\theta

so angular frequency of SHM is given as

\omega = \sqrt{\frac{mgd}{I}}

now we will have

\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}

I = 94.33 kg m^2

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The maximum speed the stuntman can have is 5.49 m/s.

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<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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