Question

A Texas cockroach of mass 0.17 kg runs counter clockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius 15 cm, rotational inertia $5.0 \times {10^{ - 3}}\;{\text{kg}} \cdot {{\text{m}}^2},$and frictionless bearings. The cockroach's speed (relative to the angular) is $2.0\;{\text{m}}/{\text{s}},$and the lazy Susan turns clockwise with angular speed$\omega = 2.8{\text{rad}}/{\text{s}}$. The Cockroach finds a bread crumb on the rim and, of course, stops. What is the angular speed of the lazy Susan after the cockroach stops?

Answer 1

Hint: In order to solve this question first we will calculate the angular momentum of the cockroach and the disk. After that we will calculate the total moment of inertia of angular momentum to get the angular speed of lazy Susan.

Formula used:
L = mvr
Where m is the cockroach's mass.
The cockroach's initial pace is v.
The lazy Susan's radius is r.

Complete step-by-step answer:
The angular speed formula is used to measure the distance covered by the body in rotations or revolutions per unit of time. The term "speed" refers to how quickly or slowly an object moves. The rotational speed of an object is measured in angular speed.

Let us assemble all the given inputs:
Let the Mass of the Texas cockroach, $m = 0.17\;{\text{kg}}$
Let the Radius of the lazy Susan, $r = 15\;{\text{cm}} = 0.15\;{\text{m}}.$
Also, Rotational inertia of the lazy Susan is given as, $I = 5 \times {10^{ - 3}}\;{\text{kg}} \cdot {{\text{m}}^2}$
Let the speed of cockroaches be (relative to the ground), $v = 2\;{\text{m}}/{\text{s}}$
The lazy Susan turns clockwise with an angular speed of , ${\omega _o} = 2.8{\text{rad}}/{\text{s}}$
The final speed of the cockroach is calculated to be, ${v_f} = 0\;{\text{m}}/{\text{s}}$
Let the initial angular momentum of the of the lazy Susan be
${L_{os}} = I{\omega _o} = 5 \times {10^{ - 3}} \times 2.8 = 0.014\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}$
Now let us find the The initial angular momentum of the cockroach about the axle of the disk
${L_{oc}} = - mvr = - 0.17 \times 2 \times 0.15 = - 0.051\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}$
Where m is the cockroach's mass.
The cockroach's initial pace is v.
The lazy Susan's radius is r.
As a result, the system's original angular momentum is equal to the sum of the cockroach's and the disk's angular momentum.
${L_o} = {L_{os}} + {L_{oc}} = 0.014 - 0.051 = - 0.037\;{\text{kg}} \cdot {{\text{m}}^2}/{\text{s}}$
Suppose After the cockroach stops, the total inertia of the spinning disk will be
${I_f} = I + m{r^2} = 5 \times {10^{ - 3}} + 0.17 \times {0.15^2} = 8.825 \times {10^{ - 3}}\;{\text{kg}} \cdot {{\text{m}}^2}$
The final angular momentum of the disk will be equal to:
${L_{fs}} = {I_f}{\omega _f} = 8.825 \times {10^{ - 3}}{\omega _f}$
Where ${\omega _f}$is the final angular velocity of the disk.
From the conservation of the total angular momentum of the system
${L_o} = {L_{fs}} + {L_{fc}}$
$ - 0.037 = 8.825 \times {10^{ - 3}}{\omega _f} + 0$
Which implies
$\therefore {\omega _f} = \dfrac{{ - 0.037}}{{8.825 \times {{10}^{ - 3}}}} = - 4.2{\text{rad}}/{\text{s}}$\[\]
Hence the angular speed of the lazy Susan after the cockroach stops is $ - 4.2{\text{rad}}/{\text{s}}$and directed in the opposite direction of the initial lazy Susan's angular speed.

Note: The moment of inertia of a solid body, also known as mass moment of inertia, angular mass, second moment of mass, or, more precisely, rotational inertia, is a quantity that calculates the torque required for a desired angular acceleration around a rotational axis, in the same way as mass determines the force required for a desired acceleration. It depends on the mass distribution of the body and the axis chosen, with greater moments necessitating more torque to adjust the rate of rotation.