Help me guys The position vector of a particle of mass2.00 kg as a function of t

Question

Help me guys The position vector of a particle of mass2.00 kg as a function of time is given by r = (6.00i + 5.00 j), where r is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time. A disk moment of inertia I_1 rotates on a frictionless vertical axle with angular speed w_L. A second disk, this one having moment of inertia I_2 and initially not rotating, drops onto the first disk. Because of friction between the surfaced, the two eventually reach the same angular speed w_f. (a) Calculate w_f (b) Calculate the ratio of the final to the initial rotational energy.

Explanation / Answer

1. r = 6i + 5j

so position is constant.

( there is no t in expression of r)

hence v = dr/dt = 0

so angular momentum will also be zero.

2 . (a) Applying angular momentum conservation,

I1 w1 + I2 x 0 = (I1 + I2) wf

wf = I1 w1 / (I1 + I2)

(b) initial KE = I1 w1^2 / 2

final KE = (I1 + I2) wf^2 /2 = (I1 + I2) I1^2 w1^1 / 2 (I1 + I2)^2

= (I1^2 / (I1 + I2)) w1^2 / 2

ratio = final / initial

= I1 / (I1 + I2)

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