A rotating uniform-density disk of radius 0.6 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 5.8 kg. A lump of clay with mass 0.5 kg falls and sticks to the outer edge of the wheel at location A,
< -0.36, 0.480, 0 > m. (Let the origin of the coordinate system be the center of the disk.) Just before the impact the clay has a speed 8 m/s, and the disk is rotating clockwise with angular speed 0.83 radians/s.
11-086-clay_ball_hits_disk.jpg
(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.)
LC,i =???kg · m2/s
(b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the center C?
LC,f = ???kg · m2/s
(c) Just after the impact, what is the angular velocity of the wheel?
omega vecf = ???radians/s
no its still just wrong
That is wrong
no its not
check the arithmetic
(a) Just before the impact, the angular momentum of the combined system of wheel plus clay about the center C, LC,i, can be calculated as the sum of the angular momentum of the wheel and the angular momentum of the clay.
The angular momentum of the wheel, Lwheel, can be calculated using the formula:
Lwheel = Iwheel * ωwheel
Where:
Iwheel is the moment of inertia of the wheel, and
ωwheel is the angular velocity of the wheel.
The moment of inertia of a uniform-density disk is given by the formula:
Iwheel = (1/2) * m * r^2
Where:
m is the mass of the wheel, and
r is the radius of the wheel.
Substituting the given values:
m = 5.8 kg
r = 0.6 m
Iwheel = (1/2) * 5.8 kg * (0.6 m)^2
Now, using the angular velocity of the wheel, ωwheel = 0.83 radians/s, we can calculate the angular momentum of the wheel:
Lwheel = Iwheel * ωwheel
Substituting the calculated value of Iwheel, we can find Lwheel.
The angular momentum of the clay, Lclay, can be calculated using the formula:
Lclay = mclay * vclay * rclay
Where:
mclay is the mass of the clay (0.5 kg),
vclay is the speed of the clay (8 m/s), and
rclay is the distance of the clay from the center C (0.36 m).
Substituting the given values, we can calculate the angular momentum of the clay.
Finally, the angular momentum of the combined system of wheel plus clay about the center C is the sum of the angular momentum of the wheel and the angular momentum of the clay:
LC,i = Lwheel + Lclay
You can substitute the calculated values to find LC,i.
(b) Just after the impact, the angular momentum of the combined system of wheel plus clay about the center C, LC,f, remains the same according to the principle of conservation of angular momentum.
LC,f = LC,i
You can substitute the calculated value of LC,i to find LC,f.
(c) Just after the impact, the angular velocity of the wheel remains the same as the initial angular velocity because the clay sticks to the outer edge of the wheel, which does not affect the distribution of mass or the moment of inertia.
So, the angular velocity of the wheel, ωvecf, remains 0.83 radians/s.
You can simply state that ωvecf = 0.83 radians/s.
To solve this problem, we need to use the principles of conservation of angular momentum. Angular momentum is defined as the product of moment of inertia and angular velocity. The total angular momentum before and after the impact should remain constant.
(a) Just before the impact, to find the angular momentum of the combined system, we need to find the angular momentum of the wheel and the angular momentum of the clay separately and then add them up.
The angular momentum of the wheel can be calculated using the formula:
Lwheel = Iwheel * ωwheel
where Iwheel is the moment of inertia of the wheel and ωwheel is the angular velocity of the wheel.
The moment of inertia of a disk rotating about its central axis is given by the formula:
Iwheel = (1/2) * m * r^2
where m is the mass of the wheel and r is the radius of the wheel.
Substituting the given values:
m = 5.8 kg
r = 0.6 m
Iwheel = (1/2) * 5.8 kg * (0.6 m)^2
The angular momentum of the wheel can be calculated using the mass of the wheel and its angular velocity:
Lwheel = (1/2) * 5.8 kg * (0.6 m)^2 * 0.83 rad/s
Next, we calculate the angular momentum of the clay. It is given by:
Lclay = mclay * r * v
where mclay is the mass of the clay, r is the distance of the clay from the center of the wheel, and v is the velocity of the clay.
Substituting the given values:
mclay = 0.5 kg
r = 0.6 m
v = 8 m/s
Lclay = 0.5 kg * 0.6 m * 8 m/s
Finally, we add the angular momenta of the wheel and the clay to find the total angular momentum just before the impact:
LC,i = Lwheel + Lclay
(b) Just after the impact, the clay sticks to the wheel, so they become a combined system. The total angular momentum should remain the same. Therefore, the angular momentum just after the impact is the same as just before the impact, i.e., LC,f = LC,i.
(c) To find the angular velocity of the wheel just after the impact, we need to calculate the moment of inertia of the combined system (wheel + clay) using the formula:
IC = Iwheel + Iclay
where Iclay is the moment of inertia of the clay.
For a point mass rotating about an axis perpendicular to its motion, the moment of inertia is given by:
Iclay = mclay * r^2
Substituting the given values:
mclay = 0.5 kg
r = 0.6 m
Iclay = 0.5 kg * (0.6 m)^2
Finally, we can find the angular velocity of the wheel just after the impact using the formula:
ωf = LC,f / IC
Substituting the calculated values for LC,f and IC, we can solve for ωf.
Remember to convert the units to kg·m^2/s for angular momentum and radians/s for angular velocity.
I (no clay) = (1/2) M R^2 = .5*5.8*.36 = 1.044
I(with clay) = 1.044 + .5(.36) = 1.224
omega initial = .83
ang momentum initial of disk = I omega =
1.044*.83 = .857
ang momentum of clay = m *r cross v
= .5 * (-.36)(8) = -1.44
sum = .857 -1.44 = -.583 kg m^2/s
that is both part (a) and part (b) because the angular momentum does not change in the collision
(c)
I w = 1.224 w = -.583
w = - .476 rad/s