A student sits on a freely rotating stool holding two weights, each of which has a mass of 3.00 kg. When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kgm2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis.
(a) Find the new angular speed of the student.
(b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.
Momentum is conserved, Figure the initial momentum. Ignore arm mass, just use the distance tot he weight and the length. Notice the radius is .5, and the mass is 6kg. Figure the moment of inertia for that.
set that equal to the final angular momentum, and solve for w.
wouldnt the radius be 1 m because the weights are 1m from the axis of rotation?
so I did
1/2(6 kg(1m)^2 and that gives me a Intertia of 3.0. I multiplied that by the angular speed given in the problem; .750
then I set that equal to 1/2(6 kg)(.300m)^2 and solved for w!
ooops, you are correct, the radisu is 1m.
im just not getting the correct answer
what could i be doing wrong?
use your weights to help you find your total inertia. so 2(3)(1)^2 +3
Which your total inertia will be 9. Now use your 9 kg * M^2 in your kinetic inertia initial. (1/2)(9)(.750)^2 which will be 2.53125. Figure out your inertia final use that to help figure out the kinetic final. don't forget Inertia inital * dist=inertia final * distance.
I = 3kgm²
m = 3kg
d1 = 1m
d2 = 0.3m
w1 = 0.75rad/s
w2 = ??
The total inertia is the sum of the individual inertias.
(I + md1² + md1²)w1 = (I + md2² + md2²)w2
w2 = (I + 2md1²)w1/(I + 2md2²) = 1.9rad/s
To find the new angular speed of the student, you need to equate the initial angular momentum to the final angular momentum. The initial angular momentum is given by the moment of inertia of the system multiplied by the initial angular speed. The final angular momentum is given by the moment of inertia of the system multiplied by the final angular speed. We can set these two equal to each other and solve for the unknown variable (final angular speed).
(a) The initial angular momentum is given by:
Initial angular momentum = initial moment of inertia × initial angular speed
= (3.00 kgm^2) × (0.750 rad/s)
The final angular momentum is given by:
Final angular momentum = final moment of inertia × final angular speed
The initial moment of inertia is the sum of the moment of inertia of the student plus stool (3.00 kgm^2) and the moment of inertia of the weights (2 × 3.00 kg × (1.00 m)^2). So, the initial moment of inertia is 9.00 kgm^2.
Using the conservation of angular momentum, we have:
(9.00 kgm^2) × (0.750 rad/s) = (9.00 kgm^2) × (final angular speed)
Simplifying the above equation:
0.750 rad/s = final angular speed
Therefore, the new angular speed of the student is 0.750 rad/s.
(b) To find the kinetic energy of the rotating system before and after the student pulls the weights inward, we can use the formulas for kinetic energy. The initial kinetic energy is given by:
Initial kinetic energy = (1/2) × initial moment of inertia × (initial angular speed)^2
= (1/2) × (9.00 kgm^2) × (0.750 rad/s)^2
The final kinetic energy is given by:
Final kinetic energy = (1/2) × final moment of inertia × (final angular speed)^2
Using the conservation of angular momentum, we know that initial moment of inertia × initial angular speed = final moment of inertia × final angular speed. Thus, we can rewrite the final kinetic energy as:
Final kinetic energy = (1/2) × (9.00 kgm^2) × (0.750 rad/s)^2
Since the initial and final angular speeds are the same, the initial and final kinetic energies will be equal.
Hence, the kinetic energy of the rotating system before pulling the weights inward is equal to the kinetic energy of the rotating system after pulling the weights inward.
To find the new angular speed of the student (part a), you need to use the principle of conservation of angular momentum. The angular momentum is given by the formula L = Iω, where L is angular momentum, I is moment of inertia, and ω is the angular speed.
In the initial state, the student has an angular momentum of LI = (3.00 kg * m^2)(0.750 rad/s) = 2.25 kg * m^2/s.
In the final state, the moment of inertia changes because the weights are pulled inward. The new moment of inertia is calculated as follows:
I_initial * r_initial = I_final * r_final
Using the initial values: (3.00 kg * m^2)(1.00 m) = I_final * (0.300 m)
I_final = (3.00 kg * m^2 * 1.00 m) / 0.300 m = 10.00 kg * m^2
Now let's use the conservation of angular momentum equation, L_initial = L_final. Rearranging the equation, we have:
I_initial * ω_initial = I_final * ω_final
Substituting the known values:
(3.00 kg * m^2)(0.750 rad/s) = (10.00 kg * m^2)(ω_final)
0.750 rad/s = (10.00 kg * m^2 / 3.00 kg * m^2)(ω_final)
0.750 rad/s = (10.00 / 3.00)(ω_final)
ω_final = (0.750 rad/s)(3.00 / 10.00)
ω_final = 0.225 rad/s
Therefore, the new angular speed of the student is 0.225 rad/s.
To find the kinetic energy of the rotating system before and after the weights are pulled inward (part b), you need to use the formula for kinetic energy of rotating bodies:
KE = (1/2) * I * ω^2
For the initial kinetic energy, substitute the known values:
KE_initial = (1/2) * (3.00 kg * m^2) * (0.750 rad/s)^2
KE_initial = 0.84375 joules
For the final kinetic energy, substitute the new moment of inertia and the new angular speed:
KE_final = (1/2) * (10.00 kg * m^2) * (0.225 rad/s)^2
KE_final = 0.253125 joules
Therefore, the kinetic energy of the rotating system before the weights are pulled inward is 0.84375 joules, and after the weights are pulled inward, it becomes 0.253125 joules.