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A thin rod of length 2.15 m and mass 13.1 kg is rotated at an angular speed of 3.95 rad/s around an axis perpendicular to the rod and through its center of mass. Find the magnitude of the rod's angular momentum.

Respuesta :

hamzaahmeds

Answer:

L = 239.2 J.s

Explanation:

The angular momentum of the rod is given by the following formula:

L = mvr

where,

L = Angular Momentum = ?

m = mass = 13.1 kg

r = radius of circle of rotation = length of rod = 2.15 m

v = linear speed of rod = r(angular speed of rod) = (2.15 m)(3.95 rad/s)

v = 8.5 m/s

Therefore,

L = (13.1 kg)(8.5 m/s)(2.15 m)

L = 239.2 J.s

shubhamchouhanvrVT

The angular momentum of the thin rod will be L = 239.2 J.s

What is angular momentum?

Whenever any object has rotational motion then its angular momentum will be the product of the linear velocity of the object and the radius of the circle of rotation of the object.

The angular momentum of the rod is given by the following formula:

[tex]L= MVr[/tex]

where,

L = Angular Momentum = ?

m = mass = 13.1 kg

r = radius of circle of rotation = length of rod = 2.15 m

w=3.95 rad/s

Now the angular momentum will be calculated as:

[tex]L= MVr[/tex]

The linear velocity will be

[tex]V=wr=3.95\times 2.15=8.5\ \frac{m}{s}[/tex]

Now put the value in the formula we get:

[tex]L=13.1\times 8.5\times 2.15=239.2\ J[/tex]

[tex]L=239.2\ J[/tex]

Hence the angular momentum of the thin rod will be L = 239.2 J.s

To know more about angular momentum follow

https://brainly.com/question/4126751

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