A student on a piano stool rotates freely with an angular speed of 2.95 rad/s . The student holds a 1.35 kg mass in each outstretched arm, 0.739 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.13 kg m^2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.51 rad/s. Having found that the new radius of rotation for the weights is .395 m, what is the initial and final kinetic energy for the system?
To find the initial and final kinetic energy of the system, we need to calculate the moment of inertia in each case and then use the formula for kinetic energy.
We are given the following information:
Initial angular speed, ω1 = 2.95 rad/s
Mass of each outstretched arm, m = 1.35 kg
Distance from the axis of rotation, r1 = 0.739 m
Combined moment of inertia, I = 5.13 kg m^2
Final angular speed, ω2 = 3.51 rad/s
New radius of rotation for the weights, r2 = 0.395 m
First, let's calculate the initial and final moments of inertia of the system:
Initial moment of inertia, I1 = I + 2(mr1)^2
= 5.13 kg m^2 + 2(1.35 kg * 0.739 m)^2
= 5.13 kg m^2 + 2(1.35 kg * 0.545521 m^2)
= 5.13 kg m^2 + 1.85 kg m^2
= 6.98 kg m^2
Final moment of inertia, I2 = I + 2(mr2)^2
= 5.13 kg m^2 + 2(1.35 kg * 0.395 m)^2
= 5.13 kg m^2 + 2(1.35 kg * 0.155025 m^2)
= 5.13 kg m^2 + 0.4185 kg m^2
= 5.55 kg m^2
Now, let's calculate the initial and final kinetic energies using the formula:
Initial kinetic energy, K1 = (1/2) * I1 * ω1^2
= (1/2) * 6.98 kg m^2 * (2.95 rad/s)^2
= (1/2) * 6.98 kg m^2 * 8.7025 rad^2/s^2
= 30.477 kg m^2/s^2
Final kinetic energy, K2 = (1/2) * I2 * ω2^2
= (1/2) * 5.55 kg m^2 * (3.51 rad/s)^2
= (1/2) * 5.55 kg m^2 * 12.3201 rad^2/s^2
= 42.985 kg m^2/s^2
Therefore, the initial kinetic energy for the system, K1, is approximately 30.477 kg m^2/s^2 and the final kinetic energy, K2, is approximately 42.985 kg m^2/s^2.