Question

021 (part 1 of 2) 10.0 points

A student sits on a rotating stool holding two
4 kg objects. When his arms are extended
horizontally, the objects are 0.7 m from the
axis of rotation, and he rotates with angular
speed of 0.69 rad/sec. The moment of iner-
tia of the student plus the stool is 4 kg mº
and is assumed to be constant. The student
then pulls the objects horizontally to a radius
0.29 m from the rotation axis.
(D.
T
TZ
Calculate the final angular speed of the
student.
Answer in units of rad/s.
022 (part 2 of 2) 10.0 points
Calculate the change in kinetic energy of the
system.
Answer in units of J.

Answer 1

Conservation of angular momentum is used to find the final angular velocity after the student pulls the weights in. To find the change in kinetic energy, the difference between the initial and final rotational kinetic energies is calculated.

Step-by-step explanation:

The student's question involves the concept of conservation of angular momentum, which states that if no external torques act on a system, the total angular momentum of the system remains constant. When the student on the rotating stool pulls the weights closer to the axis of rotation, the moment of inertia of the system decreases, causing an increase in angular velocity to conserve angular momentum. The initial angular momentum can be calculated using the initial moment of inertia and angular speed. Since the total angular momentum must remain the same, we can set it equal to the final moment of inertia (student, stool, and weights at the new radius) multiplied by the final angular speed, and solve for the final angular speed.

To calculate the change in kinetic energy, we first determine the rotational kinetic energies before and after the student pulls the weights closer using the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia and ω is the angular speed. The change in kinetic energy is the difference between the final and initial kinetic energies. This change reflects the work done by the student to pull the weights closer.

Answer 2

Answers:

a)
`1.17 rad/s`

b)
`1.31 J`

Step-by-step explanation:

a) We can solve this knowing the angular momentum
`L` is conserved, then:

`L_(o)=L_(f)`

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

Where:

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

`\omega_(o)=0.69 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_(o)` and
`I_(f)`:

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

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

Where:

`I_(s)=4 kgm^(2)` is the student's moment of inertia

`m=4 kg` is the mass of each object

`r_(o)=0.7 m` is the initial radius

`r_(f)=0.29 m` is the final radius

Then:

`I_(o)=4 kgm^(2)+2(4 kg)(0.7 m)^(2)=7.92 kgm^(2)` (4)

`I_(f)=4 kgm^(2)+2(4 kg)(0.29 m)^(2)=4.67 kgm^(2)` (5)

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

`(7.92 kgm^(2)) (0.69 rad/s)=4.67 kgm^(2)\omega_(f)` (6)

Finding
`\omega_(f)`:

`\omega_(f)=1.17 rad/s` (7) This is the final angular velocity

b) The rotational kinetic energy is:

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

The change in kinetic energy is:

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

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

`\Delta K=(1)/(2)(4.67 kgm^(2)) (1.17 rad/s)^(2)-(1)/(2)(7.92 kgm^(2)) (0.69 rad/s)^(2)` (10)

Finally:

`\Delta K=1.31 J`