Esercizi sul Baricentro in Fisica

in data
Esercizi sul Baricentro in Fisica Esercizi sul Baricentro in Fisica
Esercizi sul Baricentro in Fisica

Versione italiana

Esercizi sul Baricentro in Fisica

Concetti Chiave

Il baricentro (o centro di massa) di un sistema di punti è il punto in cui si può considerare concentrata tutta la massa del sistema. Per un sistema di punti, il baricentro GGG è dato dalla formula:

G = \left( \frac{x_1 + x_2 + \ldots + x_n}{n}, \frac{y_1 + y_2 + \ldots + y_n}{n} \right) G=(x1+x2+…+xnn,y1+y2+…+ynn) G = \left( \frac{x_1 + x_2 + \ldots + x_n}{n}, \frac{y_1 + y_2 + \ldots + y_n}{n} \right)

dove:

  • (x_i, y_i)(xi,yi)(x_i, y_i) sono le coordinate dei punti,
  • nnn è il numero totale di punti.

Baricentro di un Corpo Rigido

Per un corpo rigido, il baricentro può essere calcolato come:

G = \frac{1}{M} \sum_{i=1}^{n} m_i \cdot r_i G=1M∑i=1nmi⋅ri G = \frac{1}{M} \sum_{i=1}^{n} m_i \cdot r_i

dove:

  • MMM è la massa totale del corpo,
  • m_imim_i è la massa del iii-esimo punto,
  • r_irir_i è il vettore posizione del iii-esimo punto.

Esercizio 1: Baricentro di un Triangolo

Dati

Considera un triangolo con i vertici A(1, 2)A(1,2)A(1, 2), B(4, 6)B(4,6)B(4, 6) e C(7, 2)C(7,2)C(7, 2).

Obiettivo

Calcola le coordinate del baricentro GGG del triangolo.

Soluzione

Utilizziamo la formula del baricentro:

G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) G=(xA+xB+xC3,yA+yB+yC3) G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)

Sostituendo i valori:

G = \left( \frac{1 + 4 + 7}{3}, \frac{2 + 6 + 2}{3} \right) = \left( \frac{12}{3}, \frac{10}{3} \right) = (4, \frac{10}{3}) G=(1+4+73,2+6+23)=(123,103)=(4,103) G = \left( \frac{1 + 4 + 7}{3}, \frac{2 + 6 + 2}{3} \right) = \left( \frac{12}{3}, \frac{10}{3} \right) = (4, \frac{10}{3})

Quindi, il baricentro GGG ha coordinate (4, \frac{10}{3})(4,103)(4, \frac{10}{3}).

Esercizio 2: Baricentro di un Sistema di Punti

Dati

Considera i punti P_1(2, 3)P1(2,3)P_1(2, 3), P_2(4, 5)P2(4,5)P_2(4, 5), e P_3(6, 1)P3(6,1)P_3(6, 1).

Obiettivo

Calcola le coordinate del baricentro GGG del sistema di punti.

Soluzione

Utilizziamo la formula del baricentro:

G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) G=(x1+x2+x33,y1+y2+y33) G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Sostituendo i valori:

G = \left( \frac{2 + 4 + 6}{3}, \frac{3 + 5 + 1}{3} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3) G=(2+4+63,3+5+13)=(123,93)=(4,3) G = \left( \frac{2 + 4 + 6}{3}, \frac{3 + 5 + 1}{3} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3)

Quindi, il baricentro GGG ha coordinate (4, 3)(4,3)(4, 3).

Esercizio 3: Baricentro di un Quadrilatero

Dati

Considera un quadrilatero con i vertici A(0, 0)A(0,0)A(0, 0), B(2, 0)B(2,0)B(2, 0), C(2, 2)C(2,2)C(2, 2), e D(0, 2)D(0,2)D(0, 2).

Obiettivo

Calcola le coordinate del baricentro GGG del quadrilatero.

Soluzione

Utilizziamo la formula del baricentro:

G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4} \right) G=(xA+xB+xC+xD4,yA+yB+yC+yD4) G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4} \right)

Sostituendo i valori:

G = \left( \frac{0 + 2 + 2 + 0}{4}, \frac{0 + 0 + 2 + 2}{4} \right) = \left( \frac{4}{4}, \frac{4}{4} \right) = (1, 1) G=(0+2+2+04,0+0+2+24)=(44,44)=(1,1) G = \left( \frac{0 + 2 + 2 + 0}{4}, \frac{0 + 0 + 2 + 2}{4} \right) = \left( \frac{4}{4}, \frac{4}{4} \right) = (1, 1)

Quindi, il baricentro GGG ha coordinate (1, 1)(1,1)(1, 1).

Esercizio 4: Baricentro di un Corpo Rigido

Dati

Un corpo rigido è composto da tre masse concentrate nei seguenti punti:

  • m_1 = 2 \, \text{kg}m1=2 kgm_1 = 2 \, \text{kg} a (1, 1)(1,1)(1, 1)
  • m_2 = 3 \, \text{kg}m2=3 kgm_2 = 3 \, \text{kg} a (4, 5)(4,5)(4, 5)
  • m_3 = 5 \, \text{kg}m3=5 kgm_3 = 5 \, \text{kg} a (7, 2)(7,2)(7, 2)

Obiettivo

Calcola le coordinate del baricentro GGG del corpo rigido.

Soluzione

Calcoliamo la massa totale MMM:

M = m_1 + m_2 + m_3 = 2 \, \text{kg} + 3 \, \text{kg} + 5 \, \text{kg} = 10 \, \text{kg} M=m1+m2+m3=2 kg+3 kg+5 kg=10 kg M = m_1 + m_2 + m_3 = 2 \, \text{kg} + 3 \, \text{kg} + 5 \, \text{kg} = 10 \, \text{kg}

Ora utilizziamo la formula del baricentro:

G = \left( \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}, \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} \right) G=(m1x1+m2x2+m3x3M,m1y1+m2y2+m3y3M) G = \left( \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}, \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} \right)

Sostituendo i valori:

G = \left( \frac{2 \cdot 1 + 3 \cdot 4 + 5 \cdot 7}{10}, \frac{2 \cdot 1 + 3 \cdot 5 + 5 \cdot 2}{10} \right) G=(2⋅1+3⋅4+5⋅710,2⋅1+3⋅5+5⋅210) G = \left( \frac{2 \cdot 1 + 3 \cdot 4 + 5 \cdot 7}{10}, \frac{2 \cdot 1 + 3 \cdot 5 + 5 \cdot 2}{10} \right)

Calcoliamo le coordinate:

G_x = \frac{2 + 12 + 35}{10} = \frac{49}{10} = 4.9 Gx=2+12+3510=4910=4.9 G_x = \frac{2 + 12 + 35}{10} = \frac{49}{10} = 4.9

G_y = \frac{2 + 15 + 10}{10} = \frac{27}{10} = 2.7 Gy=2+15+1010=2710=2.7 G_y = \frac{2 + 15 + 10}{10} = \frac{27}{10} = 2.7

Quindi, il baricentro GGG ha coordinate (4.9, 2.7)(4.9,2.7)(4.9, 2.7).

Esercizio 5: Baricentro di un Sistema di Punti con Coordinate Negative

Dati

Considera i punti P_1(-3, -2)P1(−3,−2)P_1(-3, -2), P_2(-1, -4)P2(−1,−4)P_2(-1, -4), e P_3(-5, -1)P3(−5,−1)P_3(-5, -1).

Obiettivo

Calcola le coordinate del baricentro GGG del sistema di punti.

Soluzione

Utilizziamo la formula del baricentro:

G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) G=(x1+x2+x33,y1+y2+y33) G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Sostituendo i valori:

G = \left( \frac{-3 + (-1) + (-5)}{3}, \frac{-2 + (-4) + (-1)}{3} \right) = \left( \frac{-9}{3}, \frac{-7}{3} \right) = (-3, -\frac{7}{3}) G=(−3+(−1)+(−5)3,−2+(−4)+(−1)3)=(−93,−73)=(−3,−73) G = \left( \frac{-3 + (-1) + (-5)}{3}, \frac{-2 + (-4) + (-1)}{3} \right) = \left( \frac{-9}{3}, \frac{-7}{3} \right) = (-3, -\frac{7}{3})

Quindi, il baricentro GGG ha coordinate (-3, -\frac{7}{3})(−3,−73)(-3, -\frac{7}{3}).

English version

Exercises on the Center of Mass in Physics

Key Concepts

The center of mass (or center of mass) of a system of points is the point where all the mass of the system can be considered concentrated. For a system of points, the center of mass GGG is given by the formula:

G = \left( \frac{x_1 + x_2 + \ldots + x_n}{n}, \frac{y_1 + y_2 + \ldots + y_n}{n} \right) G=(x1+x2+…+xnn,y1+y2+…+ynn) G = \left( \frac{x_1 + x_2 + \ldots + x_n}{n}, \frac{y_1 + y_2 + \ldots + y_n}{n} \right)

where:

  • (x_i, y_i)(xi,yi)(x_i, y_i) are the coordinates of the points,
  • nnn is the total number of points.

Center of Mass of a Rigid Body

For a rigid body, the center of mass can be calculated as:

G = \frac{1}{M} \sum_{i=1}^{n} m_i \cdot r_i G=1M∑i=1nmi⋅ri G = \frac{1}{M} \sum_{i=1}^{n} m_i \cdot r_i

where:

  • MMM is the total mass of the body,
  • m_imim_i is the mass of the iii-th point,
  • r_irir_i is the position vector of the iii-th point.

Exercise 1: Center of Mass of a Triangle

Data

Consider a triangle with vertices A(1, 2)A(1,2)A(1, 2), B(4, 6)B(4,6)B(4, 6) and C(7, 2)C(7,2)C(7, 2).

Objective

Calculate the coordinates of the center of mass GGG of the triangle.

Solution

We use the centroid formula:

G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) G=(xA+xB+xC3,yA+yB+yC3) G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)

Substituting the values:

G = \left( \frac{1 + 4 + 7}{3}, \frac{2 + 6 + 2}{3} \right) = \left( \frac{12}{3}, \frac{10}{3} \right) = (4, \frac{10}{3}) G=(1+4+73,2+6+23)=(123,103)=(4,103) G = \left( \frac{1 + 4 + 7}{3}, \frac{2 + 6 + 2}{3} \right) = \left( \frac{12}{3}, \frac{10}{3} \right) = (4, \frac{10}{3})

So, the centroid GGG has coordinates (4, \frac{10}{3})(4,103)(4, \frac{10}{3}).

Exercise 2: Center of Mass of a System of Points

Data

Consider the points P_1(2, 3)P1(2,3)P_1(2, 3), P_2(4, 5)P2(4,5)P_2(4, 5), and P_3(6, 1)P3(6,1)P_3(6, 1).

Objective

Calculate the coordinates of the center of mass GGG of the system of points.

Solution

We use the centroid formula:

G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) G=(x1+x2+x33,y1+y2+y33) G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Substituting the values:

G = \left( \frac{2 + 4 + 6}{3}, \frac{3 + 5 + 1}{3} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3) G=(2+4+63,3+5+13)=(123,93)=(4,3) G = \left( \frac{2 + 4 + 6}{3}, \frac{3 + 5 + 1}{3} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3)

So, the centroid GGG has coordinates (4, 3)(4,3)(4, 3).

Exercise 3: Centroid of a Quadrilateral

Data

Consider a quadrilateral with vertices A(0, 0)A(0,0)A(0, 0), B(2, 0)B(2,0)B(2, 0), C(2, 2)C(2,2)C(2, 2), and D(0, 2)D(0,2)D(0, 2).

Objective

Calculate the coordinates of the centroid GGG of the quadrilateral.

Solution

We use the centroid formula:

G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4} \right) G=(xA+xB+xC+xD4,yA+yB+yC+yD4) G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4} \right)

Substituting the values:

G = \left( \frac{0 + 2 + 2 + 0}{4}, \frac{0 + 0 + 2 + 2}{4} \right) = \left( \frac{4}{4}, \frac{4}{4} \right) = (1, 1) G=(0+2+2+04,0+0+2+24)=(44,44)=(1,1) G = \left( \frac{0 + 2 + 2 + 0}{4}, \frac{0 + 0 + 2 + 2}{4} \right) = \left( \frac{4}{4}, \frac{4}{4} \right) = (1, 1)

So, the centroid GGG has coordinates (1, 1)(1,1)(1, 1).

Exercise 4: Center of Mass of a Rigid Body

Data

A rigid body is composed of three masses concentrated in the following points:

  • m_1 = 2 \, \text{kg}m1=2 kgm_1 = 2 \, \text{kg} at (1, 1)(1,1)(1, 1)
  • m_2 = 3 \, \text{kg}m2=3 kgm_2 = 3 \, \text{kg} at (4, 5)(4,5)(4, 5)
  • m_3 = 5 \, \text{kg}m3=5 kgm_3 = 5 \, \text{kg} at (7, 2)(7,2)(7, 2)

Objective

Calculate the coordinates of the center of mass GGG of the rigid body.

Solution Let's calculate the total mass MMM:

M = m_1 + m_2 + m_3 = 2 \, \text{kg} + 3 \, \text{kg} + 5 \, \text{kg} = 10 \, \text{kg}M=m1+m2+m3=2 kg+3 kg+5 kg=10 kgM = m_1 + m_2 + m_3 = 2 \, \text{kg} + 3 \, \text{kg} + 5 \, \text{kg} = 10 \, \text{kg}
Now we use the center of gravity formula:
G = \left( \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}, \frac{m_1 y_ 1 + m_2 y_2 + m_3 y_3}{M} \right)G=(m1x1+m2x2+m3x3M,m1y1+m2y2+m3y3M)G = \left( \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}, \frac{m_1 y_ 1 + m_2 y_2 + m_3 y_3}{M} \right)
Substituting the values:
G = \left( \frac{2 \cdot 1 + 3 \cdot 4 + 5 \cdot 7}{10}, \frac{2 \cdot 1 + 3 \cdot 5 + 5 \cdot 2}{10} \right) G=(2⋅1+3⋅4+5⋅710,2⋅1+3⋅5+5⋅210)G = \left( \frac{2 \cdot 1 + 3 \cdot 4 + 5 \cdot 7}{10}, \frac{2 \cdot 1 + 3 \cdot 5 + 5 \cdot 2}{10} \right)

Let's calculate the coordinates:

G_x = \frac{2 + 12 + 35}{10} = \frac{49}{10} = 4.9 Gx=2+12+3510=4910=4.9 G_x = \frac{2 + 12 + 35}{10} = \frac{49}{10} = 4.9

G_y = \frac{2 + 15 + 10}{10} = \frac{27}{10} = 2.7 Gy=2+15+1010=2710=2.7 G_y = \frac{2 + 15 + 10}{10} = \frac{27}{10} = 2.7

So, the centroid GGG has coordinates (4.9, 2.7)(4.9,2.7)(4.9, 2.7).

Exercise 5: Centroid of a System of Points with Negative Coordinates

Data

Consider the points P_1(-3, -2)P1(−3,−2)P_1(-3, -2), P_2(-1, -4)P2(−1,−4)P_2(-1, -4), and P_3(-5, -1)P3(−5,−1)P_3(-5, -1).

Objective

Calculate the coordinates of the centroid GGG of the system of points.

Solution

We use the centroid formula:

G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) G=(x1+x2+x33,y1+y2+y33) G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Substituting the values:

G = \left( \frac{-3 + (-1) + (-5)}{3}, \frac{-2 + (-4) + (-1)}{3} \right) = \left( \frac{-9}{3}, \frac{-7}{3} \right) = (-3, -\frac{7}{3}) G=(−3+(−1)+(−5)3,−2+(−4)+(−1)3)=(−93,−73)=(−3,−73) G = \left( \frac{-3 + (-1) + (-5)}{3}, \frac{-2 + (-4) + (-1)}{3} \right) = \left( \frac{-9}{3}, \frac{-7}{3} \right) = (-3, -\frac{7}{3})

So, the centroid GGG has coordinates (-3, -\frac{7}{3})(−3,−73)(-3, -\frac{7}{3}).

Commenti

Posta un commento