Suppose a friend is completing another physics assignment in their chair at thei

Question

Suppose a friend is completing another physics assignment in their chair at their desk and you take the opportunity to turn this scenario into the physics question you now must complete. Your friend and chair have a combined constant moment of inertia of 4.6 kg m2, and the chair can freely rotate. Now your friend picks up two solid bricks (each of mass 1.4 kg) they accidentally collected from one of the campus buildings, and experiment with rotational kinematics. Supposing they hold the masses in each hand, arms outstretched, at distance of 0.7 m from their axis of rotation, and have an initial angular speed of 2.7 rev/s.

a) Calculate the distance in m of each brick from your friend’s axis of rotation if they pull the bricks inward such that their rotational speed has increased to 3.2 rev/s.

Answer: 0.396

b) What is the initial rotational kinetic energy of the system?

Answer: 21.76794

c) What is the final rotational kinetic energy of the system?

Answer: 25.79904

Could you please write down the full solution?

Explanation / Answer

Moment of inertia of each Brick = Mx2
M is mass of each brick and x is distance of Bricks from axis of rotation.
Moment of inertia of system with Bricks at distance of 0.7 m from axis
I1 = 4.6 + 2*1.4*0.72 = 5.972 Kg m2
Moment of inertia of system with Bricks at distance 'x' m from axis, after pulled closer to axis,
I2 = 4.6 + 2*1.4*x2 = 4.6 + 2.8x2
As Bricks are pulled closer to axis, angular momentum (Iw) remains constant. (as there there is no external torque on the system)
hence I1 w1 = I2 w2 ( w = 2 pi f , f is frequency of rotation)
I1 2 pi f1 = I2 2 pi f2
I1 f1 = I2 f2
5.972*2.7 = (4.6+2.8x2)* 3.2
x = 0.396 m

Rotational Kinetic energy = Iw2/2
b) Initial Rotational kinetic energy = 5.972*(2 pi 2.7)2/2
                                          = 858.49 J   ( answer given is wrong. w is taken as 2.7, while it is
                                                                frequency of rotation, not angular velocity)

c) Final Rotational kinetic energy = (4.6 + 2.8*0.3962)(2 pi 2.7)2/2
                                                  = 1017.51 J

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