Question

6_1. A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm.

a. Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. (Part A figure) What is its angular momentum? Express your answer in kilogram meters squared per second.

6_2. Learning Goal: To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies.
Consider a turntable to be a circular disk of moment of inertia I(sub(t)) rotating at a constant angular velocity omega(sub(i)) around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I(sub(r)) . The initial angular velocity of the second disk is zero. There is friction between the two disks.
After this "rotational collision," the disks will eventually rotate with the same angular velocity.a.What is the final angular velocity, omega(sub(f)), of the two disks? Express omega(sub(f)) in terms of I(sub(t)), I(sub(r)), and omega(sub(i)).
omega(sub(f)) =?
b.Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, K(f), of the two spinning disks? Express the final kinetic energy in terms of I(sub(t)), I(sub(r)), and the initial kinetic energy K(i) of the two-disk system. No angular velocities should appear in your answer.
K(f) =?
c.Assume that the turntable deccelerated during time delta(t) before reaching the final angular velocity (delta(t)is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, (tau), acting on the bottom disk due to friction with the record? Express the torque in terms of I(sub(t)), omega(sub(i)), omega(sub(f)), and delta(t).
Tau =?

Answer 1

Concepts and reason

This question is based on the rotational motion of an object.

Initially, find the moment of inertia of the object and then substitute in its expression.

For the final kinetic energy and final angular velocity, apply law of conservation of momentum. For the average torque use its expression.

Fundamentals

Consider a rod rotating of mass m rotating with angular velocity $\omega$ .

The moment of inertia of the rod about an axis perpendicular to it, passing through the center of the rod is,

$I = \frac{1}{{12}}m{l^2}$

Here, l is the length of the rod.

The angular momentum L of the rod is,

$L = I\omega$

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

As per law of conservation of angular momentum, the angular momentum of a system remains constant unless acted on by an external torque.

The rotational kinetic energy K of an object is,

$K = \frac{1}{2}I{\omega ^2}$

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

The expression for rotational torque is given as,

$\tau = I\frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{{\Delta t}}$

Here, ${\omega _{\rm{f}}}$ and ${\omega _{\rm{i}}}$ are the final and initial angular velocity of the object and $\Delta t$ is the small time in which the change take place.

(6.1.a)

Consider a baton of mass m rotating with angular velocity $\omega$ .

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

$L = \frac{1}{{12}}m{l^2}\omega$

Here, l is the length of the baton.

Substitute $0.120{\rm{ kg}}$ for m, ${\rm{80}}{\rm{.0 cm}}$ for l and ${\rm{3}}{\rm{.00 rad/s}}$ ,

$\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}$

(6.2.a)

Consider a system of two disks. Let the disk falls on another disk rotating with angular velocity ${\omega _{\rm{i}}}$ .After that, the two disks start rotating with angular velocity ${\omega _{\rm{f}}}$ .

As the second disk is initially at rest, so the initial angular momentum ${L_{_{\rm{i}}}}$ of the system is,

${L_{_{\rm{i}}}} = {I_{\rm{t}}}{\omega _{\rm{i}}}$

Here, ${I_{\rm{t}}}$ is the moment of inertia of the rotating disk.

The final angular momentum of the system is the sum of the angular momentum of the two disks and given as follows:

${L_{_{\rm{f}}}} = \left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right){\omega _{\rm{f}}}$

Here ${I_{\rm{r}}}$ is the moment of inertia of the second disk and ${\omega _{\rm{f}}}$ is the final angular velocity with which the disks rotate.

Apply the law of conservation of angular momentum as follows:

$\begin{array}{c}\\{L_{_{\rm{f}}}} = {L_{_{\rm{i}}}}\\\\\left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right){\omega _{\rm{f}}} = {I_{\rm{t}}}{\omega _{\rm{i}}}\\\\{\omega _{\rm{f}}} = \frac{{{I_{\rm{t}}}{\omega _{\rm{i}}}}}{{\left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right)}}\\\end{array}$

(6.2.b)

Find the final rotational kinetic energy of the two spinning disks.

The initial kinetic energy ${K_{\rm{i}}}$ of the system is,

${K_{\rm{i}}} = \frac{1}{2}{I_{\rm{t}}}{\omega ^2}_{\rm{i}}$

The final rotational kinetic energy ${K_{\rm{f}}}$ of the system is the sum of the kinetic energy of the two disks.

${K_{\rm{f}}} = \frac{1}{2}\left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right){\omega ^2}_{\rm{f}}$

Substitute the expression for final angular velocity ${\omega _{\rm{f}}} = \frac{{{I_{\rm{t}}}{\omega _{\rm{i}}}}}{{\left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right)}}$ ’

$\begin{array}{c}\\{K_f} = \frac{1}{2}\left( {{I_{\rm{t}}} + {I_{\rm{r}}}} \right){\left( {\frac{{{I_{\rm{t}}}{\omega _{\rm{i}}}}}{{{I_{\rm{t}}} + {I_{\rm{r}}}}}} \right)^2}\\\\ = \frac{1}{2}\frac{{{{\left( {{I_{\rm{t}}}{\omega _{\rm{i}}}} \right)}^2}}}{{{I_{\rm{t}}} + {I_{\rm{r}}}}}\\\end{array}$

Substitute the values from the expression of initial kinetic energy as follows:

${K_{\rm{i}}} = \frac{1}{2}{I_{\rm{t}}}{\omega ^2}_{\rm{i}}$

$\begin{array}{c}\\{K_{\rm{f}}} = \frac{1}{2}{I_{\rm{t}}}\omega _{\rm{i}}^2\frac{{{I_{\rm{t}}}}}{{{I_{\rm{t}}} + {I_{\rm{r}}}}}\\\\ = {K_{\rm{i}}}\frac{{{I_{\rm{t}}}}}{{{I_{\rm{t}}} + {I_{\rm{r}}}}}\\\end{array}$

(6.2.c)

The torque $\tau$ acting on the disk is given as follows:

$\tau = {I_{\rm{t}}}\alpha$

Here, $\alpha$ is the angular acceleration of the object and given as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{{\Delta t}}$

Here, $\Delta t$ is the time interval in which the velocity of the moving disk changes from ${\omega _{\rm{i}}}$ to ${\omega _{\rm{f}}}$ .

So, the average torque is given as follows:

$\tau = {I_{\rm{t}}}\left( {\frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{{\Delta t}}} \right)$

Ans: Part 6.1.a

The angular momentum of the baton is $0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}$ .