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Over [174]
2 years ago
11

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

its center, perpendicular to the plate. Express your answer in terms of the variables l, w, and M.
Physics
1 answer:
Ad libitum [116K]2 years ago
4 0

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

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Explanation:

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The tilt of the Earth's axis of rotation changes from approximately 22.5° to 24.5° over a period of 41,000 years. This causes gl
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What is the acceleration of a boy on a skateboard if the net force on the boy is 15 N? The total mass of the boy and the skatebo
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                                 Force = (mass) x (acceleration)

If the full 15N is pointing parallel to the ground,
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                                   15 N  =  (58 kg) x (acceleration).

Divide each side
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                                                         = (15 kg-m/s²) / (58 kg)

                                                         = (15/58) (kg-m/kg-s²)

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What would be the best design for an experiment that tests how much water expands when frozen?
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B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately  about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

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5 0
1 year ago
A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1m/s^2 for 20s.
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Answer:

A.) 1430 metres

B.) 80 seconds

Explanation:

Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:

U = acceleration × time

U = 1.1 × 20 = 22 m/s

It then proceeds at constant speed for 1100 m

Then, time t will be

Time = distance/ velocity

Time = 1100/22

Time = 50 s

before slowing down at 2.2m/s^2 until it stops at the station.

Deceleration = velocity/time

2.2 = 22/t

t = 22/2.2

t = 10s

Using area under the graph, the distance between the two stations will be :

(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)

220 + 1100 + 110

1430 m

The time taken between the two stations will be

20 + 50 + 10 = 80 seconds

6 0
3 years ago
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