Question

Learning Goal: To learn the definition andapplications of angular momentum including its relationship totorque.

By now, you should be familiar with the concept ofmomentum, defined as the product of an object's mass andits velocity:

$\vec p = m\vec v$ .

You may have noticed that nearly every translationalconcept or equation seems to have an analogous rotationalone. So, what might be the rotational analogue of momentum?

Just as the rotational analogue of force $F_vec$ , called thetorque $tau_vec$ , isdefined by the formula

$\vec{\tau} = \vec{r} \times \vec{F}$ ,

the rotational analogue of momentum $p_vec$ , called theangular momentum $L_vec$ , is givenby the formula

$\vec{L} = \vec{r} \times \vec{p}$ ,

for a single particle. For an extended body you must add up theangular momenta of all of the pieces.

There is another formula for angular momentum that makes theanalogy to momentum particularly clear. For a rigid body rotatingabout an axis of symmetry, which will be true for all parts in thisproblem, the measure of inertia is given not by the mass but by therotational inertia (i.e., the moment of inertia). Similarly, therate of rotation is given by the body's angular speed,. Theproduct gives the angular momentum $L_vec$ of a rigidbody rotating about an axis of symmetry. (Note that if the body isnot rotating about an axis of symmetry, then the angular momentumand the angular velocity may not be parallel.)

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Part A

Which of the following is the SI unit ofangular momentum?

$\rm N\cdot m/s$ $\rm kg \cdot m/s$ $\rm kg \cdot m^2/s^2$ $\rm kg \cdot m^2/s$

Part B

An object has rotational inertia . The object,initially at rest, begins to rotate with a constant angularacceleration of magnitude . What isthe magnitude of the angular momentum ofthe object after time ?

Express your answer in terms of, , and.

=

...............................................

Part C

A rigid, uniform bar with mass and length rotates aboutthe axis passing through the midpoint of the bar perpendicular tothe bar. The linear speed of the end points of the bar is. What is themagnitude of the angular momentum ofthe bar?

Express your answer in terms of, , , andappropriate constants.

=

Answer 1

Concepts and reason

The concepts required to solve the given questions angular momentum and the moment of inertia of the rod.

Initially, calculate the SI unit of angular momentum. Later, calculate the magnitude of the angular momentum. Finally, calculate the momentum of inertia of rod and the angular momentum.

Fundamentals

The expression for the angular momentum is as follows:

$\vec L = \vec r \times \vec p$

Here, $\vec L$ is the angular momentum, $\vec r$ is the position vector, and $\vec p$ is the linear momentum.

The expression for the magnitude of the angular momentum in terms of the moment of inertia and angular velocity is as follows:

$\vec L = \vec I \times \vec \omega$

Here, $\vec I$ is the moment of inertia and $\vec \omega$ is the angular velocity.

The expression for the angular speed in terms of the acceleration magnitude with time t is as follows:

$\omega = \alpha t$

Here, $\alpha$ is the angular acceleration.

Thus, the expression for the angular momentum is as follows:

$L = I\alpha t$

The expression for the formula of moment of inertia of the rod is as follows:

$I = \int\limits_0^M {{r^2}} dm$

Here, r is the perpendicular distance and dm is the elementary mass.

The expression for the angular velocity in terms of the radius and the speed is as follows:

$\omega = \frac{v}{r}$

Here, v is the speed and r is the radius.

(A)

Substitute ${\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for $\vec p$, ${\rm{m}}$ for $\vec r$ in the equation $\vec L = \vec r \times \vec p$.

$\begin{array}{c}\\\vec L = {\rm{m}} \times {\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\\\\ = {\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

(B)

The expression for the angular momentum is given by,

$\vec L = \vec I \times \vec \omega$

Substitute $\alpha t$ for $\vec \omega$ in the equation $\vec L = \vec I \times \vec \omega$.

$L = I\alpha t$

(C)

Calculate the moment of inertia of the bar using the equation $I = \int\limits_0^M {{r^2}} dm$ as follows:

$\begin{array}{c}\\I = \int\limits_0^M {{r^2}\left( {\frac{M}{L}} \right)} dm\\\\ = \frac{M}{L}\left( {\frac{{{r^3}}}{3}} \right)_{ - L/2}^{L/2}\\\\ = \frac{1}{{12}}M{L^2}\\\end{array}$

Thus, the rotational inertia of the bar is equal to $\frac{1}{{12}}M{L^2}$.

Now, solve for the angular speed by substituting $\frac{b}{2}$ for r in the equation $\omega = \frac{v}{r}$.

$\begin{array}{c}\\\omega = \frac{v}{{\frac{b}{2}}}\\\\ = \frac{{2v}}{b}\\\end{array}$

Thus, the angular speed of the bar is $\frac{{2v}}{b}$.

Now, substitute L for b in the equation $I = \frac{1}{{12}}M{L^2}$ ,

$I = \frac{1}{{12}}M{L^2}$

Finally, substitute $\frac{1}{{12}}M{b^2}$ for I and $\frac{{2v}}{b}$ for $\vec \omega$ in the equation $\vec L = \vec I \times \vec \omega$.

$\begin{array}{c}\\L = \left( {\frac{1}{{12}}M{b^2}} \right)\left( {\frac{{2v}}{b}} \right)\\\\ = \frac{{Mbv}}{6}\\\end{array}$

Ans: Part A

The SI unit of angular momentum is equal to ${\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}$.