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Leokris [45]
3 years ago
9

2. An object with moment of inertia ????1 = 9.7 x 10−4 kg ∙ m2 rotates at a speed of 3.0 rev/s. A 20 g mass with moment of inert

ia ????2 = 1.32 x 10−6 kg ∙ m2 is dropped onto the rotating object at a distance of 5.0 cm from the center of mass. What is the angular velocity of the combined object and mass after the drop?
Physics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

20 g = 0.02 kg

When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

I_2 = 1.32\times10^{−6} + 0.02 * 0.05^2 = 0.513\times10^{-4} kgm^2

So the total moment of inertia of the system of 2 objects after the drop is:

I = I_1 + I_2 = 9.7\times10^{-4} + 0.513\times10^{-4} = 0.0010213 kgm^2

From here we can apply the law of angular momentum conservation to calculate the post angular speed

\omega_1 I_1 = \omega_2 I

\omega_2 = \omega_1 \frac{I_1}{I} = 3 \frac{9.7\times10^{-4}}{0.0010213} = 2.85 rad/s

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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