# Moment of I, Angular momentum, K

• joemama69
In summary, the moment of inertia of the ruler is 1/12 of its mass plus (.5x + .5l), the angular momentum is 2Ipi2/P2, and the torque is -mghsinQ.
joemama69

## Homework Statement

A ruler of mass m, length l, and width h has a hole drilled into it a short distance x from one end and equidistant from both sides, as shown in the diagram. It is anchored on a frictionless air table by a nail driven through the hole and is then set rotating about the nail. For the following questions, the answers to any part can include the symbols representing the answers to previous parts, whether or not the previous parts were correctly answered. There is no need to substitute the expressions from the previous parts.

a) Find the moment of inertia I of the ruler about the axis of rotation.

b) If the rotation has a period P, what is the magnitude L of the angular momentum of the rotating ruler?

c) Continuing to take P as the period of rotation, what is the kinetic energy K of the ruler?

The ruler is now suspended by a fixed horizontal rod, which passes through the hole drilled in the ruler and forms a frictionless pivot. The ruler is held so that it makes an angle θ to the vertical, as shown in the
diagram, and is then released. Take the acceleration of gravity as g (g > 0).

d) Just after it is released, what is the torque acting on
it? Since the rotation is about a fixed axis, we can use the scalar
definition of torque, with the positive direction taken as counter-
clockwise. (If you prefer to describe the torque as a vector, that
would also be completely acceptable.)

## The Attempt at a Solution

part a) i used the parralele axis theorem

I = Icm + Mh2 = 1/12 M (l2 + h2) + M(.5l - x)2

I realize i can simplify but is this correct so far

part b)

is it just L = Iw where it is the I from part a and then w = (2pi)/P

part c)

K = .5Iw2 where I is from part a and w is from part b so...K = 2Ipi2/P2

part d) still working on

#### Attachments

• 3b.pdf
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• 3.pdf
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joemama69 said:
part a) i used the parralele axis theorem

I = Icm + Mh2 = 1/12 M (l2 + h2) + M(.5l - x)2

I realize i can simplify but is this correct so far

Yes

is it just L = Iw where it is the I from part a and then w = (2pi)/P

Yes
K = .5Iw2 where I is from part a and w is from part b so...K = 2Ipi2/P2

Yes.

Have more confidence in yourself!

thanks,

part d)

is it just Toruqe = -mghsinQ

do u do small approx an dmake it simQ=Q

## 1. What is the Moment of Inertia?

The Moment of Inertia is a measure of an object's resistance to changes in its rotational motion. It is represented by the symbol "I" and is dependent on the mass distribution of the object and its rotational axis.

## 2. How is the Moment of Inertia calculated?

The Moment of Inertia is calculated by multiplying the mass of each particle in an object by the square of its distance from the rotational axis and then summing up these values for all particles in the object. The formula is I = ∑miri2.

## 3. What is Angular Momentum?

Angular Momentum is a measure of an object's rotational motion. It is represented by the symbol "L" and is dependent on the object's moment of inertia, angular velocity, and direction of rotation. It is a conserved quantity, meaning it remains constant unless acted upon by an external torque.

## 4. How is Angular Momentum related to Moment of Inertia?

Angular Momentum and Moment of Inertia are directly related, as they both involve an object's rotational motion. The moment of inertia determines how difficult it is to change an object's rotational motion, while angular momentum measures the amount of rotational motion an object has. An object with a larger moment of inertia will have a larger angular momentum at the same angular velocity.

## 5. What is the relationship between Angular Momentum and Kinetic Energy?

Angular Momentum and Kinetic Energy are both measures of an object's motion, but they are not directly related. Kinetic Energy is dependent on an object's mass and linear velocity, while Angular Momentum is dependent on an object's moment of inertia and angular velocity. However, in a closed system with no external torques acting, the total angular momentum and total kinetic energy will remain constant.

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