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Oksi-84 [34.3K]

3 years ago

15

Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,

5.00) m, and a 3.40-kg object at (2.00, 0) m. Where should a fourth object of mass 6.00 kg be placed so that the center of gravity of the four-object arrangement will be at (0, 0)?

1 answer:

adelina 88 [10]3 years ago

7 0

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

x_cm = 1 /M ∑ $x_{i}$ m_{i}

y_cm = 1 /M ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

x4=?

y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

M = 2.00 + 2.20 + 3.40

M = 7.6 kg

0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

x4 = -6.8 / 6

x4 = -1,133 m

Axis y

0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

y4 = -11/6

y4 = -1.83 m

body position 4 is (-1,133, -1.83)

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Explanation:

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Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

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