Question

A thin rod of length 3.05 m and mass 15.7 kg is rotated at an angular speed of 3.60 rad/s around an axis perpendicular to the rod and through one of its ends. Find the magnitude of the rod's angular momentum.

Answer 1

Answer:

526 kg-m2/sec

Explanation:

The radius = length of the rod= 3.05 m

angular velocity = 3.60 rad/s

mass = 15.7 kg

L = I ω

Where;

L = angular momentum

I = moment of inertia = mr^2

ω = angular velocity

substituting;

L= 15.7 * (3.05)^2 * 3.60

L = 526 kg-m2/sec

Answer 2

The magnitude of the rod's angular momentum will be 526 Kgm²/sec.

What is angular momentum.?

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The given data in the problem is;

r is the radius of the rod =3.05 m

l is the length of rod  =3.05 m

$\rm \omega$ is the angular velocity = 3.60 rad/s

m is the mass = 15.7 kg

The angular momentum is given by;

$\rm L=I \omega \\\\ \rm L=mr^ 2 \times \omega \\\\ \rm L=15.07 \times (3.05)^2 \times 3.60 \\\\ \rm L=526 \ Kg m^2 /sec$

Hence the magnitude of the rod's angular momentum will be 526 Kgm²/sec.

To learn more about the angular momentum refer to the link;

https://brainly.com/question/15104254