A professor sits on a rotating stool that is spinning at 10.010.0 rpm while she holds a heavy weight in each of her hands. Her outstretched hands are 0.7850.785 m from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to 32.532.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in?

Question

Coleman Brock

Physics

15 May, 07:21

A professor sits on a rotating stool that is spinning at 10.010.0 rpm while she holds a heavy weight in each of her hands. Her outstretched hands are 0.7850.785 m from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to 32.532.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in?

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Answers (2)

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Gabriel Jacobson · Answer 1

0.44 m

Explanation:

w₀ = initial angular speed of rotating stool = 10.0 rpm

r₀ = initial distance of weights from axis of rotation = 0.785 m

w = final angular speed = 32.5 rpm

r = final distance of the weights from axis of rotation = ?

m = mass of weights

Using conservation of angular momentum

m r₀² w₀ = m r² w

(0.785) ² (10) = r² (32.5)

r = 0.44 m

Lilian Young · Answer 2

0.44 m

Explanation:

f = 10 rpm = 10 / 60 rps

ω = 2 x 3.14 x 10 / 60 = 1.047 rad/s

f' = 32.5 rpm = 32.5 / 60 rps

ω' = 2 x 3.14 x 32.5 / 60 = 3.4 rad/s

r = 0.785 m

Let the mass of each block be m and the new distance is r'.

As no external torque is there, so angular momentum remains constant

I x ω = I' x ω'

2 m r^2 x ω = 2 m r'^2 x ω'^2

0.785 x 0.785 x 1.047 = r'^2 x 3.4

r'^2 = 0.1898

r' = 0.44 m