A student on a piano stool rotates freely with an angular speed of 3.07 rev/s. The student holds a 1.38 kg mass in each outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 4.72 kg·m2, a value that remains constant.
a. As the student pulls his arms inward, his angular speed increases to 3.54 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?
b. Calculate the initial and final kinetic energy of the system.
This is a conservation of angular momentum question.
Li = I1 omegai + I2 omegai
I1 given
I2 = mr^2 times 2 (since 2 books)
omegai is the same for both (convert to rad/sec)
Lf = I1 omegaf + I2 omegaf
compute new I2
Li=Lf and solve for omegaf
b) KEi = 1/2I1 omegai^2 + 1/2I2 omegai^2
do same for KEf
a. To solve this problem, we need to apply the conservation of angular momentum. According to this principle, the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the student is pulling his arms inward, changing the distribution of mass and consequently the moment of inertia.
The formula for angular momentum is given by:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Initially, the angular momentum of the system is given by:
L1 = I1 * ω1
Final angular momentum is given by:
L2 = I2 * ω2
Since the combined moment of inertia of the student and the stool remains constant, we can write:
L1 = L2
Let's substitute the values given:
I1 * ω1 = I2 * ω2
Rearranging the equation to solve for I2:
I2 = (I1 * ω1) / ω2
Now we can calculate I2 using the given values:
I2 = (4.72 kg·m^2 * 3.07 rev/s) / 3.54 rev/s
Simplifying the equation:
I2 = 4.09 kg·m^2
To find the distance of the masses from the axis of rotation, we can use the formula for moment of inertia of point masses:
I = m * r^2
Where m is the mass and r is the radius (distance from the axis of rotation).
Rearranging the equation, we can solve for r:
r = sqrt(I / m)
Substituting the values:
r = sqrt(4.09 kg·m^2 / 1.38 kg)
Simplifying the equation:
r = 1.46 m
Therefore, the masses are 1.46 meters from the axis of rotation at this time.
b. To calculate the initial and final kinetic energy of the system, we can use the formula:
KE = 1/2 * I * ω^2
Let's calculate the initial kinetic energy:
KE1 = 1/2 * (4.72 kg·m^2 + 2 * 1.38 kg * (0.759 m)^2) * (3.07 rev/s)^2
Simplifying the equation:
KE1 = 206.27 Joules
Now, let's calculate the final kinetic energy:
KE2 = 1/2 * (4.72 kg·m^2 + 2 * 1.38 kg * (1.46 m)^2) * (3.54 rev/s)^2
Simplifying the equation:
KE2 = 269.26 Joules
Therefore, the initial kinetic energy of the system is 206.27 Joules, and the final kinetic energy is 269.26 Joules.