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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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Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

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3 years ago
How is lithium fluoride and lithium chloride similar?
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Answer:

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Explanation:

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3 years ago
What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake
Sav [38]

Answer:

Controlled braking

Explanation:

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3 years ago
Only kealani57 can have it no one can have this points if you take the points you will get reported
MatroZZZ [7]

Hi sorry she is a little mean sorry

:(        :(        :(       :(      :(     :(      :(      :(     :(     :(     :(     :(     :(     :(     :(     :(     :(

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0.035 N

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6 0
3 years ago
Read 2 more answers
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