Question

PART ONE

A student sits on a rotating stool holding two
2 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.73 rad/sec. The moment of inertia of the
student plus the stool is 8 kg m2
and is assumed to be constant. The student then pulls
the objects horizontally to a radius 0.28 m
from the rotation axis.
Calculate the final angular speed of the
student.
Answer in units of rad/s.
(USE PICTURES)

PART TWO
Calculate the change in kinetic energy of the
system.
Answer in units of J.

Answer 1

To find the final angular speed of the rotating student who pulls objects closer, we apply the conservation of angular momentum, which allows us to solve for the new angular speed. To calculate the change in kinetic energy, we compare the initial and final kinetic energies of the system, considering the changes in moment of inertia and angular speed.

Step-by-step explanation:

The question involves the concept of conservation of angular momentum and relates to the change in angular speed when a student on a rotating stool moves two 2 kg objects closer to the axis of rotation. Initially, the objects are 1 m away from the rotation axis and the system rotates with an angular speed of 0.73 rad/sec. The moment of inertia (I) of the student plus the stool is 8 kg·m2, and the student pulls the objects to a radius of 0.28 m.

Using conservation of angular momentum, we can say that the initial angular momentum (Linitial) equals the final angular momentum (Lfinal). The formula for angular momentum is L = I·ω, where ω is the angular speed. Therefore:

Linitial = (I + 2·m·rinitial2)·ωinitial

Lfinal = (I + 2·m·rfinal2)·ωfinal

Plugging in the values:

8 kg·m2 + 2(2 kg)(1 m)2)· 0.73 rad/s = (8 kg·m2 + 2(2 kg)(0.28 m)2)·ωfinal

Solving for ωfinal, we find the final angular speed of the student.

For the change in kinetic energy (KE), we calculate the initial and final kinetic energies using KE = (1/2)Iω2, and then find the difference between them to get the change in kinetic energy.

Answer 2

Answers:

a)
`1.05 rad/s`

b)
`1.38 J`

Step-by-step explanation:

a) Final angular velocity :

Before solving this part, we have to stay clear that the angular momentum
`L` is conserved, since we are dealing with circular motion, then:

`L_(o)=L_(f)`

Hence:

`I_(i) \omega_(i)=I_(f) \omega_(f)` (1)

Where:

`I_(i)` is the initial moment of inertia of the system

`\omega_(i)=0.73 rad/s` is the initial angular velocity

`I_(f)` is the final moment of inertia of the system

`\omega_(f)` is the final angular velocity

But first, we have to find
`I_(i)` and
`I_(f)`:

`I_(i)=I_(s)+2mr_(i)^(2)` (2)

`I_(f)=I_(s)+2mr_(f)^(2)` (3)

Where:

`I_(s)=8 kgm^(2)` is the student's moment of inertia

`m=2 kg` is the mass of each object

`r_(i)=1 m` is the initial radius

`r_(f)=0.28 m` is the final radius

Then:

`I_(i)=8 kgm^(2)+2(2 kg)(1 m)^(2)=12 kgm^(2)` (4)

`I_(f)=8 kgm^(2)+2(2 kg)(0.28 m)^(2)=8.31 kgm^(2)` (5)

Substituting the results of (4) and (5) in (1):

`(12 kgm^(2)) (0.73 rad/s)=8.31 kgm^(2)\omega_(f)` (6)

Finding
`\omega_(f)`:

`\omega_(f)=1.05 rad/s` (7) This is the final angular speed

b) Change in kinetic energy:

The rotational kinetic energy is defined as:

`K=(1)/(2)I \omega^(2)` (8)

And the change in kinetic energy is:

`\Delta K=(1)/(2)I_(f) \omega_(f)^(2)-(1)/(2)I_(i) \omega_(i)^(2)` (9)

Since we already calculated these values, we can solve (9):

`\Delta K=(1)/(2)(8.31 kgm^(2)) (1.05 rad/s)^(2)-(1)/(2)(12 kgm^(2)) (0.73 rad/s)^(2)` (10)

Finally:

`\Delta K=1.38 J` This is the change in kinetic energy