A rotating uniform-density disk of radius 0.7 m is mounted in the vertical plane. The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 5 kg. A lump of clay with mass 0.3 kg falls and sticks to the outer edge of the wheel at location A, < -0.455, 0.532, 0 > m. (Let the origin of the coordinate system be the center of the disk.) Just before the impact the clay has a speed 7 m/s, and the disk is rotating clockwise with angular speed 0.23 radians/s.
(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C? (As usual, x is to the right, y is up, and z is out of the screen, toward you.)
b. final angular momentum
c. Just after the impact, what is the angular velocity of the wheel?
To answer these questions, we need to analyze the conservation of angular momentum. The angular momentum equation states that the initial angular momentum of a system is equal to the final angular momentum of the system, assuming no external torques act on the system.
(a) To find the initial angular momentum of the system just before the impact of the clay, we need to calculate the angular momentum of the wheel and the linear momentum of the clay.
The angular momentum of the wheel is given by the formula:
L_wheel = I * ω_wheel
where I is the moment of inertia of the wheel and ω_wheel is the angular velocity of the wheel.
The moment of inertia of a uniform-density disk rotating about its axis is given by:
I_disk = (1/2) * m * r^2
where m is the mass of the disk and r is the radius of the disk.
Plugging in the given values, we have:
I_disk = (1/2) * 5 kg * (0.7 m)^2
= 0.6125 kg·m^2
The angular momentum of the wheel is then:
L_wheel = I_disk * ω_wheel
= 0.6125 kg·m^2 * 0.23 rad/s
= 0.140875 kg·m^2/s
Next, we calculate the linear momentum of the clay.
Linear momentum, p, is given by:
p = m * v
where m is the mass of the clay and v is its velocity.
The linear momentum of the clay just before the impact is:
p_clay = m_clay * v_clay
= 0.3 kg * 7 m/s
= 2.1 kg·m/s
Now, the initial angular momentum of the combined system of the wheel and clay is given by the sum of the angular momentum of the wheel and the linear momentum of the clay:
L_initial = L_wheel + p_clay
Plugging in the values, we get:
L_initial = 0.140875 kg·m^2/s + 2.1 kg·m/s
= 2.240875 kg·m^2/s
Therefore, the angular momentum of the combined system just before the impact is 2.240875 kg·m^2/s.
(b) To find the final angular momentum of the system, we need to consider the sticking of the clay to the wheel. Since the clay sticks to the wheel, it becomes part of the rotating system.
The final angular momentum of the system after the clay sticks to the wheel is simply the angular momentum of the wheel and the clay about the center C:
L_final = I_final * ω_final
Since the clay sticks to the outer edge, the moment of inertia of the system (wheel + clay) about the center C is:
I_final = I_wheel + I_clay
The moment of inertia of a particle of mass m at a distance r from an axis is given by:
I_particle = m * r^2
Plugging in the values, we get:
I_clay = m_clay * r_clay^2
= 0.3 kg * (-0.455 m)^2
= 0.06140825 kg·m^2
Therefore, the moment of inertia of the system (wheel + clay) about the center C is:
I_final = I_disk + I_clay
= 0.6125 kg·m^2 + 0.06140825 kg·m^2
= 0.67390825 kg·m^2
Now, we can calculate the final angular momentum of the system:
L_final = I_final * ω_final
= 0.67390825 kg·m^2 * ω_final
Unfortunately, we don't have the value of the final angular velocity (ω_final) given in the problem statement, so we cannot calculate the final angular momentum without it.
(c) Similarly, we cannot determine the angular velocity of the wheel just after the impact without knowing the final angular momentum or any other related information.