A student holds two lead weights, each of mass
12 kg. When the students’ arms are extended
horizontally, the lead weights are 0.82 m from
the axis of rotation and the student rotates
with an angular speed of 1.5 rad/sec. The
moment of inertia of student plus stool is
5.5 kg m2
and is assumed to be constant; i.e.,
the student’s arms are massless! Then the
student pulls the lead weights horizontally to
a radius 0.16 m from the axis of rotation.
Calculate the final angular speed of the
system.
Answer in units of rad/s
We can solve this problem using the conservation of angular momentum. The initial angular momentum (L1) is equal to the final angular momentum (L2).
L1 = L2
Angular momentum is the product of moment of inertia (I) and angular velocity (ω).
L1 = I1 * ω1
L2 = I2 * ω2
where I1 and I2 are the moments of inertia before and after the student pulls the weights in, and ω1 and ω2 are the angular velocities before and after, respectively.
Initially, the student has a moment of inertia of 5.5 kg m^2, and each lead weight has a moment of inertia equal to its mass (12 kg) times the square of the distance to the axis of rotation (0.82 m).
I_lead = 2*(12 kg * (0.82 m)²) = 16.2336 kg m^2.
So the total moment of inertia initially is:
I1 = 5.5 kg m^2 + 16.2336 kg m^2 = 21.7336 kg m^2.
After the student pulls the weights, the new moment of inertia of the lead weights becomes:
I_lead_new = 2*(12 kg * (0.16 m)²) = 0.6144 kg m^2.
So the total moment of inertia after the student pulls the weights is:
I2 = 5.5 kg m^2 + 0.6144 kg m^2 = 6.1144 kg m^2.
Now we can apply the conservation of angular momentum.
I1 * ω1 = I2 * ω2
21.7336 kg m^2 * 1.5 rad/s = 6.1144 kg m^2 * ω2
Solving for ω2:
ω2 = (21.7336 kg m^2 * 1.5 rad/s) / 6.1144 kg m^2 = 5.3189 rad/s.
The final angular speed of the system is 5.3189 rad/s.
To calculate the final angular speed of the system, we can use the principle of conservation of angular momentum. The angular momentum before the student pulls the lead weights can be equated to the angular momentum after the weights are pulled.
The initial angular momentum is given by:
L_initial = I_initial * ω_initial
Where:
L_initial = Initial angular momentum
I_initial = Initial moment of inertia (student plus stool)
ω_initial = Initial angular speed
The final angular momentum is given by:
L_final = I_final * ω_final
Where:
L_final = Final angular momentum
I_final = Moment of inertia after weights are pulled
ω_final = Final angular speed
Since the moment of inertia includes the masses of the lead weights, we need to find the moment of inertia of the lead weights at both initial and final positions.
The moment of inertia of a point mass rotating about an axis is given by:
I = m * r^2
Where:
I = Moment of inertia
m = Mass of the point mass
r = Distance from the axis of rotation to the point mass
The moment of inertia before pulling the weights is:
I_initial = moment of inertia of student plus stool + 2 * moment of inertia of lead weights
I_initial = 5.5 kg m^2 + 2 * (12 kg * (0.82 m)^2)
I_initial = 5.5 kg m^2 + 2 * 95.616 kg m^2
I_initial = 5.5 kg m^2 + 191.232 kg m^2
I_initial = 196.732 kg m^2
The moment of inertia after pulling the weights is:
I_final = moment of inertia of student plus stool + 2 * moment of inertia of lead weights
I_final = 5.5 kg m^2 + 2 * (12 kg * (0.16 m)^2)
I_final = 5.5 kg m^2 + 2 * 3.072 kg m^2
I_final = 5.5 kg m^2 + 6.144 kg m^2
I_final = 11.644 kg m^2
Since angular momentum is conserved, we can equate the initial and final angular momenta:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Solving for ω_final:
ω_final = (I_initial * ω_initial) / I_final
Substituting the values:
ω_final = (196.732 kg m^2 * 1.5 rad/s) / 11.644 kg m^2
ω_final = 295.098 kg m^2 rad/s / 11.644 kg m^2
ω_final ≈ 25.34 rad/s
Therefore, the final angular speed of the system is approximately 25.34 rad/s.
To calculate the final angular speed of the system, we can make use of the principle of conservation of angular momentum.
The initial angular momentum of the system is equal to the final angular momentum of the system. Mathematically, this can be expressed as:
Initial angular momentum = Final angular momentum
The initial angular momentum can be calculated as the product of the moment of inertia and the initial angular speed:
Initial angular momentum = moment of inertia * initial angular speed
The final angular momentum can be calculated as the product of the moment of inertia and the final angular speed:
Final angular momentum = moment of inertia * final angular speed
Since the moment of inertia of the student plus stool is assumed to be constant, we can set up the equation:
moment of inertia * initial angular speed = moment of inertia * final angular speed
Now, we can solve for the final angular speed:
final angular speed = (moment of inertia * initial angular speed) / moment of inertia
To solve the equation, substitute the given values:
moment of inertia = 5.5 kg m^2
initial angular speed = 1.5 rad/s
final angular speed = (5.5 kg m^2 * 1.5 rad/s) / 5.5 kg m^2
Simplifying the equation:
final angular speed = 1.5 rad/s
Therefore, the final angular speed of the system is 1.5 rad/s.