I AM SORRY I JUST HAVE ONE OF THE MOST HORRIBLE TEACHERS, AND I AM TEACHING MYSELF FOR I GET NOTHING IN CLASS.
On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.565m from the axis of rotation of the stool. She is given an angular velocity of 2.80rad/s , after which she pulls the dumbbells in until they are only 0.200m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.90kg⋅m2 and may be considered constant. Each dumbbell has a mass of 4.95kg and may be considered a point mass. Neglect friction.
Part A
What is the initial angular momentum of the system?
ANSWER IN kg⋅m2/s in the direction of the motion
Part B
What is the angular velocity of the system after the dumbbells are pulled in toward the axis?
ω =
rad/s
Part C
Compute the kinetic energy of the system before the dumbbells are pulled in.
K = J
Part D
Compute the kinetic energy of the system after the dumbbells are pulled in.K = J
I initial = 4.90 + .565^2 (4.95*2)
= 4.90 + .319 (9.9)
= 4.90 + 3.16
= 8.06
Ifinal = 4.90 + .2^2(9.9)
= 4.90 + .396
= 5.30 oh my, much less
angular momentum initial and final
= I omega1 = 8.06*2.8
= 22.6 (answer A)
= 5.3*omega2
so
omega 2 = 22.6/5.3 = 4.26 rad/s
Now I think you can do
(1/2) I omega^2 for both
You will find the Ke is higher after she pulls the dumb bells in. She has to put work in to pull the weights in because they are held out by centripetal acceleration.
Thanks! you are a life saver
Part A: "Hmm, let me calculate the initial angular momentum of the system. Doing some quick math... Ah, here it is! The initial angular momentum of the system is <insert answer> kg⋅m2/s in the direction of the motion. Make sure to use your right hand for the direction!"
Part B: "Now, let's find out the angular velocity of the system after the dumbbells are pulled in. Hold on... crunching some numbers... Voila! The angular velocity of the system is <insert answer> rad/s. Keep on spinning!"
Part C: "Time for some kinetic energy calculations! Before the dumbbells are pulled in, the system's kinetic energy is <insert answer> J. That's energy in action!"
Part D: "And finally, after the dumbbells are pulled in, we need to figure out the new kinetic energy of the system. Drum roll, please... Ta-da! The kinetic energy of the system is <insert answer> J. Keep that energy flowing!"
To find the answers to the questions, we need to apply certain formulas and principles from the field of physics.
Part A: Initial Angular Momentum
Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω):
L = I * ω
Given:
I = 4.90 kg⋅m^2 (moment of inertia)
ω = 2.80 rad/s (angular velocity)
Using the given values, we can calculate the initial angular momentum (L):
L = 4.90 kg⋅m^2 * 2.80 rad/s
L = 13.72 kg⋅m^2/s
The initial angular momentum of the system is 13.72 kg⋅m^2/s in the direction of the motion.
Part B: Angular Velocity after Dumbbells are Pulled In
According to the Law of Conservation of Angular Momentum, the total angular momentum of an isolated system remains constant if no external torques act on it. Therefore, the initial angular momentum in Part A remains constant.
L_initial = L_final
Given:
L_initial = 13.72 kg⋅m^2/s
After the dumbbells are pulled in, the moment of inertia decreases. The new moment of inertia (I_final) can be calculated using the parallel axis theorem:
I_final = I_cm + 2 * m * r^2
where I_cm is the moment of inertia about the center of mass, m is the mass of each dumbbell, and r is the new distance from the axis of rotation.
Given:
m = 4.95 kg (mass of each dumbbell)
r = 0.200 m (new distance from the axis)
Calculating I_final:
I_cm = m * r^2 = 4.95 kg * (0.200 m)^2 = 0.198 kg⋅m^2
I_final = 0.198 kg⋅m^2 + 2 * (4.95 kg) * (0.200 m)^2
I_final = 0.198 kg⋅m^2 + 0.198 kg⋅m^2 = 0.396 kg⋅m^2
Now, using the conservation of angular momentum equation, we can find the final angular velocity (ω_final):
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Substituting the known values:
4.90 kg⋅m^2 * 2.80 rad/s = 0.396 kg⋅m^2 * ω_final
Solving for ω_final:
ω_final = (4.90 kg⋅m^2 * 2.80 rad/s) / (0.396 kg⋅m^2)
ω_final = 34.65 rad/s
The angular velocity of the system after the dumbbells are pulled in is 34.65 rad/s.
Part C: Initial Kinetic Energy
The formula for kinetic energy (K) is given by:
K = (1/2) * I * ω^2
Given:
I = 4.90 kg⋅m^2 (moment of inertia)
ω = 2.80 rad/s (angular velocity)
Using the given values, we can calculate the initial kinetic energy (K):
K = (1/2) * 4.90 kg⋅m^2 * (2.80 rad/s)^2
K = 19.488 J (Joules)
The kinetic energy of the system before the dumbbells are pulled in is 19.488 Joules.
Part D: Final Kinetic Energy
Using the final angular velocity (ω_final) from Part B, we can calculate the final kinetic energy (K_final) using the same formula as in Part C:
K_final = (1/2) * I_final * ω_final^2
Given:
I_final = 0.396 kg⋅m^2 (new moment of inertia)
ω_final = 34.65 rad/s (final angular velocity)
Calculating K_final:
K_final = (1/2) * 0.396 kg⋅m^2 * (34.65 rad/s)^2
K_final = 247.219 Joules
The kinetic energy of the system after the dumbbells are pulled in is 247.219 Joules.