Question

A student on a piano stool rotates freely with an angular speed of 2.97 rev/s. The student holds a 1.32-kg mass in each outstretched arm, 0.797 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.35 kgm2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.46 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points

Answer 1

Distance of masses from the axis of rotation is now; 0.508 m

Step-by-step explanation:

We are given;

Initial angular velocity;ω_i = 2.97 rev/s

Final angular velocity;ω_f = 3.46 rev/s

Mass;m = 1.32 kg

Initial distance;r_i = 0.797 m

Combined moment of inertia of student and stool; I_c = 5.35 kgm²

Now, from conservation of angular momentum, we know that;

Initial angular momentum(L_i) = Final angular momentun(L_f)

Thus;L_i = L_f

So, we have; (I_i)•ω_i = (I_f)•ω_f

I_i would be calculated from;

I_i = I_c + 2m(r_i)²

So, I_i = 5.35 + 2(1.32*0.796²)

I_i = 7.023 kg.m²

So,from (I_i)•ω_i = (I_f)•ω_f, plugging in the relevant values, we have;

7.023 x 2.97 = (I_f) x 3.46

(I_f) = (7.023 * 2.97)/3.46

(I_f) = 6.03 kg.m²

Now we'll calculate the final distance(r_f) from;

I_f = I_c + 2m(r_f)²

Plugging in the relevant values, we'll obtain;

6.03 = 5.35 + 2*1.32*(r_f)²

2*1.32*(r_f)² = 6.03 - 5.35

2.64*(r_f)² = 0.68

(r_f)² = 0.68/2.64

r_f = √0.257576

r_f = 0.508 m