Answer :
Answer:
(a)[tex]\omega_f=1.7892\,rad.s{-1}[/tex]
(b)
before [tex]KE_i=1.7297\,J[/tex]
after [tex]KE_f=5.5018\,J[/tex]
Explanation:
Given:
mass of each object, [tex]m=2.6\, kg[/tex]
initial stretch of arms from the center, [tex]r_i=1\,m[/tex]
initial angular velocity, [tex]\omega_i=0.75\,rad .s^{-1}[/tex]
moment of inertia of stool and the student together, [tex]I=3\,kg.m^2[/tex]
final stretch of arms from the center, [tex]r_{f} =0.29\,m[/tex]
According to the law of conservation of angular momentum, initial angular momentum is equal to the final angular momentum in absence of any external force.
[tex]I_i.\omega_i=I_f.\omega_f[/tex]
(a)
[tex][3+2\times 2.6\times (1)^2]\times 0.75=[3+2\times 2.6\times (0.29)^2]\times \omega_f[/tex]
[tex]\omega_f=1.7892\,rad.s{-1}[/tex]
(b)
We know,
[tex]KE= \frac{1}{2} I.\omega^2[/tex]
Putting the initial values:
Kinetic energy before the objects are pulled in
[tex]KE_i=\frac{1}{2} [3+2\times 2.6\times (1)^2]\times 0.75^2[/tex]
[tex]KE_i=1.7297\,J[/tex]
Putting the final values:
Kinetic energy after the objects are pulled in
[tex]KE_f=\frac{1}{2} [3+2\times 2.6\times (0.29)^2]\times 1.7892^2[/tex]
[tex]KE_f=5.5018\,J[/tex]
a. The new angular speed of the student in rad/s is 1.79 rad/s.
b. The kinetic energy of the student before and after the objects are pulled in, are 2.31 Joules and 5.51 Joules respectively.
Given the following data:
- Mass of object = 2.6 kg
- Initial radius = 1.0 m
- Initial angular speed = 0.75 rad/s
- Moment of inertia (student + stool) = 3.0 [tex]kgm^2[/tex]
- Final radius = 0.29 m
a. To find the new angular speed of the student in rad/s, we would use the law of conservation of angular momentum:
Applying the law of conservation of angular momentum:
[tex]I_i\omega_i = I_f\omega_f[/tex] ....equation 1
Where:
- is the initial moment of inertia.
- is the initial angular speed.
- is the final moment of inertia.
- is the final angular speed.
For initial moment of inertia:
[tex]I_i = 3 + (2[2.6] \times 1^2)\\\\I_i = 3 +5.2\\\\I_i = 8.2 \;kgms[/tex]
For final moment of inertia:
[tex]I_f = 3 + (2[2.6] \times 0.29^2)\\\\I_f = 3 +0.44\\\\I_f = 3.44 \;kgms[/tex]
Substituting the parameters into eqn 1, we have;
[tex]8.2 \times 0.75 = 3.44 \times\omega_f\\\\6.15 = 3.44\omega_f\\\\\omega_f = \frac{6.15}{3.44} \\\\\omega_f = 1.79 \; rad/s[/tex]
b. To find the kinetic energy of the student before and after the objects are pulled in:
The kinetic energy before:
[tex]K.E_i =\frac{1}{2} I_i\omega_i^2\\\\K.E_i = \frac{1}{2}\times 8.2 \times 0.75^2\\\\K.E_i = 4.1\times0.5625\\\\K.E_i =2.31 \;Joules[/tex]
The kinetic energy after:
[tex]K.E_f =\frac{1}{2} I_f\omega_f^2\\\\K.E_f = \frac{1}{2}\times 3.44 \times 1.79^2\\\\K.E_f = 1.72\times3.2041\\\\K.E_f =5.51\; Joules[/tex]
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