A student sits on a rotating stool holding two 3.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is
3.0 kg · m2
and is assumed to be constant. The student then pulls in the objects horizontally to 0.43 m from the rotation axis.
(a) Find the new angular speed of the student.
rad/s
(b) Find the kinetic energy of the student before and after the objects are pulled in.
before J
after J
To find the new angular speed of the student, we can use the principle of conservation of angular momentum.
The angular momentum of the student before pulling in the objects is given by:
L1 = I1 * ω1
where:
L1 is the initial angular momentum
I1 is the moment of inertia of the student plus stool (given as 3.0 kg · m^2)
ω1 is the initial angular speed (given as 0.75 rad/s)
The angular momentum of the student after pulling in the objects is given by:
L2 = I2 * ω2
where:
L2 is the final angular momentum
I2 is the moment of inertia of the student plus stool with the objects pulled in (not given)
ω2 is the final angular speed (to be determined)
According to the conservation of angular momentum, L1 = L2.
So, I1 * ω1 = I2 * ω2
Now, we need to find the moment of inertia of the student plus stool with the objects pulled in.
The moment of inertia of the student plus stool with the objects extended is given by:
I1 = Istudent + Iobjects
where:
Istudent is the moment of inertia of the student plus stool without any objects (not given)
Iobjects is the moment of inertia of the two 3.8 kg objects with a distance of 1.0 m from the rotation axis
The moment of inertia of two objects rotating in a circle about an axis at a distance R is given by:
I = 2 * m * R^2
where:
m is the mass of each object (3.8 kg)
R is the distance of each object from the rotation axis (1.0 m)
So, Iobjects = 2 * 3.8 kg * (1.0 m)^2 = 7.6 kg · m^2
Now, we can substitute the values into our equation:
3.0 kg · m^2 * 0.75 rad/s = (Istudent + 7.6 kg · m^2) * ω2
We want to find ω2, so we isolate it:
ω2 = (3.0 kg · m^2 * 0.75 rad/s) / (Istudent + 7.6 kg · m^2)
Now, let's move on to find the kinetic energy before and after pulling in the objects.
The kinetic energy of the student before pulling in the objects is given by:
KE1 = (1/2) * I1 * ω1^2
where:
KE1 is the initial kinetic energy (to be determined)
I1 is the moment of inertia of the student plus stool (given as 3.0 kg · m^2)
ω1 is the initial angular speed (given as 0.75 rad/s)
Substituting the values:
KE1 = (1/2) * 3.0 kg · m^2 * (0.75 rad/s)^2
Similarly, the kinetic energy of the student after pulling in the objects is given by:
KE2 = (1/2) * I2 * ω2^2
where:
KE2 is the final kinetic energy (to be determined)
I2 is the moment of inertia of the student plus stool with the objects pulled in (not given)
ω2 is the final angular speed (to be determined)
Substituting the values and simplifying:
KE2 = (1/2) * (Istudent + 7.6 kg · m^2) * ω2^2
Now, we can go back to our formula for ω2 and substitute it into the equation for KE2:
KE2 = (1/2) * (Istudent + 7.6 kg · m^2) * [(3.0 kg · m^2 * 0.75 rad/s) / (Istudent + 7.6 kg · m^2)]^2
Simplifying this expression will give the final answer for the kinetic energy after pulling in the objects.
Note: Without the value of the moment of inertia of the student plus stool (Istudent), we cannot provide the exact values for the new angular speed (rad/s), initial kinetic energy (J), and final kinetic energy (J). These calculations require the specific moment of inertia, which is not given in the question.