Respuesta :
Answer:
a
The New angular speed is [tex]w_f = 2.034 rad/s[/tex]
b
The Kinetic energy before the masses are pulled in is [tex]KE_i = 3.101 \ J[/tex]
c
The Kinetic energy after the masses are pulled in is [tex]KE_f = 8.192 \ J[/tex]
Explanation:
From the we are told that masses are 1.08 m from the axis of rotation, this means that
The radius [tex]r =1.08m[/tex]
The mass is [tex]m = 3.09\ kg[/tex]
The angular speed [tex]w = 0.770 \ rad/sec[/tex]
The moment of inertia of the system excluding the two mass [tex]I = 3.25 \ kg \cdot m^2[/tex]
New radius [tex]r_{new} = 0.34m[/tex]
Generally the conservation of angular momentum can be mathematical represented as
[tex]w_f = [\frac{I_i}{I_f} ]w_i .....(1)[/tex]
Where [tex]w_f[/tex] is the final angular speed
[tex]w_i[/tex] is the initial angular speed
[tex]I_i[/tex] is the initial moment of inertia
[tex]I_f[/tex] is the final moment of inertia
Moment of inertia is mathematically represented as
[tex]I = m r^2[/tex]
Where I is the moment of inertia
m is the mass
r is the radius
So the Initial moment of inertia is given as
[tex]I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2[/tex]
[tex]I_i = 2m r^2 + 3.25[/tex]
The multiplication by is because we are considering two masses
[tex]I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2[/tex]
So the final moment of inertia is given as
[tex]I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2[/tex]
Substituting these values into equation 1
[tex]w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec[/tex]
Generally Kinetic energy is mathematically represented in term of moment of inertia as
[tex]KE = \frac{1}{2} * I * w^2[/tex]
Now considering the kinetic energy before the masses are pulled in,
[tex]KE_i = \frac{1}{2} * I_i * w^2_i[/tex]
The Moment of inertia would be [tex]I_i = 10.46 \ Kg \cdot m^2[/tex]
The Angular speed would be [tex]w_i = 0.77 \ rad/s[/tex]
Now substituting these value into the equation above
[tex]KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J[/tex]
Now considering the kinetic energy after the masses are pulled in,
[tex]KE_f = \frac{1}{2} * I_f * w^2_f[/tex]
The Moment of inertia would be [tex]I_f = 3.96 \ Kg \cdot m^2[/tex]
The Angular speed would be [tex]w_f = 2.034 \ rad/s[/tex]
Now substituting these value into the equation above
[tex]KE_f= \frac{1}{2} *(3.96)(2.034)^2[/tex]
[tex]= 8.192J[/tex]