A student sits on a rotating stool holding two 3.4-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 0.45 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)
forgot the 1/2 in the last 2 parts
Use conservation of momentum
Initial momentum:
The momentum of inertia of the objects the students is m*r^2
(3.0 + 3.4*1^2 + 3.4*1^2)*wi = (3.0 + 3.4*.45^2 + 3.4*.45^2)*wf
where wi = 0.75 is the initial angular momentum, and wf is the final angular momentum. Solve for wf
b) Kinetic energy = 1/2*I*w^2
Initial kinetic energy is
(3.0 + 3.4*1^2 + 3.4*1^2)*wi^2
Final kinetic energy is
(3.0 + 3.4*.45^2 + 3.4*.45^2)*wf^2
To solve this problem, we can use the law of conservation of angular momentum which states that the angular momentum of an object remains constant as long as there are no external torques acting on it.
The initial angular momentum of the system is given by:
L_initial = (moment of inertia of student + stool) * angular speed_initial
(a) To find the new angular speed of the student after pulling in the objects, we can equate the initial and final angular momenta:
L_initial = L_final
(moment of inertia of student + stool) * angular speed_initial = (moment of inertia of student + stool) * angular speed_final
Since the moment of inertia of the student plus stool is assumed to be constant, we can simplify the equation as follows:
angular speed_initial = angular speed_final
Therefore, the new angular speed of the student after pulling in the objects is the same as the initial angular speed, which is 0.75 rad/s.
(b) The kinetic energy of an object rotating about an axis is given by:
KE = (1/2) * (moment of inertia) * (angular speed)^2
The initial kinetic energy of the student is:
KE_initial = (1/2) * (3.0 kg * m^2) * (0.75 rad/s)^2
KE_initial = 0.84375 J
The final kinetic energy of the student after pulling in the objects is:
KE_final = (1/2) * (3.0 kg * m^2) * (0.75 rad/s)^2
KE_final = 0.84375 J
Therefore, the kinetic energy of the student remains the same before and after pulling in the objects, which is 0.84375 J.
To find the new angular speed of the student, we can use the principle of conservation of angular momentum. Angular momentum is defined as the product of moment of inertia and angular speed, and is conserved when no external torques are acting on the system.
The initial angular momentum of the student and the objects can be calculated using the formula for angular momentum:
L_initial = I_initial * ω_initial
where
L_initial is the initial angular momentum
I_initial is the initial moment of inertia (student plus stool)
ω_initial is the initial angular speed
Given:
I_initial = 3.0 kg · m^2
ω_initial = 0.75 rad/s
From the given information, we can calculate the initial angular momentum:
L_initial = (3.0 kg · m^2) * (0.75 rad/s)
= 2.25 kg · m^2/s
Next, we need to consider the conservation of angular momentum when the student pulls the objects in. The final moment of inertia will be different from the initial value because the objects are pulled in closer to the rotation axis. Let's call the final moment of inertia I_final and the final angular speed ω_final.
Since angular momentum is conserved, we can write:
L_initial = L_final
Substituting the expressions for angular momentum:
I_initial * ω_initial = I_final * ω_final
We are given the initial values for I_initial and ω_initial, but we need to find out the final values for I_final and ω_final.
For the final moment of inertia, we can use the conservation of mass. The total mass of the student and the objects remains constant:
m_initial = m_final
where
m_initial is the initial mass of the system (student + objects)
m_final is the final mass of the system (student without the objects)
The initial system mass is the sum of the student mass (m_student) and the mass of the two objects (2 * m_object). Let's call the mass of each object m_object.
m_initial = m_student + 2 * m_object
Since the student is pulling in the objects horizontally, no external torques are acting on the system, implying that the angular momentum has to be conserved. So,
m_initial * r_initial * ω_initial = m_final * r_final * ω_final
where
r_initial is the initial distance of the objects from the axis of rotation
r_final is the final distance of the objects from the axis of rotation
We are given the initial value of r_initial as 1.0 m. To find the final value of r_final, we use the relationship:
m_initial * r_initial = m_final * r_final
Substituting the expression for m_initial:
(m_student + 2 * m_object) * r_initial = m_final * r_final
Now we have two equations to solve:
I_initial * ω_initial = I_final * ω_final
(m_student + 2 * m_object) * r_initial = m_final * r_final
From the first equation, we can solve for I_final:
I_final = (I_initial * ω_initial) / ω_final
From the second equation, we can solve for r_final:
r_final = ((m_student + 2 * m_object) * r_initial) / m_final
Substituting the expressions for I_initial, ω_initial, r_initial, and m_final:
I_final = ((3.0 kg · m^2) * (0.75 rad/s)) / ω_final
r_final = (((m_student + 2 * m_object) * 1.0 m)) / m_final
Now we have expressions for I_final and r_final in terms of ω_final and m_final. We can substitute these expressions into the equation for angular momentum conservation:
(I_initial * ω_initial) = (I_final * ω_final)
(m_student + 2 * m_object) * r_initial = m_final * r_final
Solving these equations simultaneously will give us the values of ω_final and m_final.
Once we have the value of ω_final, we can calculate the kinetic energy before and after the objects are pulled in.
The initial kinetic energy (K_initial) can be calculated as:
K_initial = (1/2) * I_initial * ω_initial^2
The final kinetic energy (K_final) can be calculated as:
K_final = (1/2) * I_final * ω_final^2
Substituting the values of I_initial, I_final, ω_initial, and ω_final will give us the answers for part (b) of the question.
Note that the mass of the student (m_student) and the mass of each object (m_object) are not given, so we would need additional information to calculate these values and arrive at the final answers.