Chemistry, 21.01.2020 19:31 raquelqueengucci25

Two thin rods of length l are rotating with the same angular speed ω (in rad/s) about axes that pass perpendicularly through one end. rod a is massless but has a particle of mass 0.76 kg attached to its free end. rod b has a mass of 0.76 kg, which is distributed uniformly along its length. the length of each rod is 0.73 m, and the angular speed is 4.7 rad/s. find the kinetic energies of rod a with its attached particle and of rod b.

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Answer from: lashondrascott

Explanation:

For rod A:

Length = 0.73 m

Mass of particle attached, m = 0.76 kg

Moment of inertia of this system about the given axis is as follows.

I = $ml^{2}$

= $0.76 kg \times (0.73)^{2}$

= 0.405 $kg m^{2}$

Angular speed of the rod ( $\omega$ ) = 4.7 rad/s

Now, the relation between kinetic energy, moment of inertia and angular speed is as follows.

K.E = $\frac{1}{2} I \omega^{2}$

= $\frac{1}{2} 0.405 kg m^{2} (4.7)^{2}$

= 4.47 J

Therefore, kinetic energy of rod A is 4.47 J.

For rod B:

Length = 0.73 m

Mass of the rod, m = 0.76 kg

Moment of inertia will be as follows.

I = $\frac{1}{3} ml^{2}$

Putting the given values into the above formula as follows.

I = $\frac{1}{3} ml^{2}$

= $\frac{1}{3} \times 0.76 kg (0.73)^{2}$

= 0.135 $kg m^{2}$

Angular speed is given as 4.7 rad/s. Therefore, calculate the value of kinetic energy as follows.

K.E = $\frac{1}{2} I \omega^{2}$

= $\frac{1}{2} \times 0.135 \times (5.7)^{2}$

= 2.19 J

Hence, kinetic energy of rod B is 2.19 J.

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