This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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The center of mass (CM) is the unique point where the weighted relative position of the distributed mass sums to zero. It's the average position of all the mass in a system. The center of gravity (CG) is the point where the entire weight of an object appears to act. For objects in a uniform gravitational field, the center of mass and center of gravity are at the same location. Here are the steps and examples for finding the center of mass: Formulas for Center of Mass For a system of discrete particles with masses m_1, m_2, , m_n located at positions (x_1, y_1), (x_2, y_2), , (x_n, y_n): x = (m_1 x_1 + m_2 x_2 + + m_n x_n)/(m_1 + m_2 + + m_n) = _i=1^n m_i x_i_i=1^n m_i y = (m_1 y_1 + m_2 y_2 + + m_n y_n)/(m_1 + m_2 + + m_n) = _i=1^n m_i y_i_i=1^n m_i For a composite lamina (a flat, thin object of uniform density), the mass is proportional to its area. So, we can replace mass (m) with area (A) in the formulas, where A_i is the area of each part and (x_i, y_i) is the centroid of that part: x = _i=1^n A_i x_i_i=1^n A_i y = _i=1^n A_i y_i_i=1^n A_i --- Example 1: Discrete Particles Find the center of mass of three particles with masses 2\,kg, 3\,kg, and 5\,kg located at coordinates (1, 2), (4, 1), and (2, 5) respectively. Step 1: Identify the masses and their coordinates. • m_1 = 2\,kg at (x_1, y_1) = (1, 2) • m_2 = 3\,kg at (x_2, y_2) = (4, 1) • m_3 = 5\,kg at (x_3, y_3) = (2, 5) Step 2: Calculate the sum of the masses. m_i = m_1 + m_2 + m_3 = 2\,kg + 3\,kg + 5\,kg = 10\,kg Step 3: Calculate the sum of the products of mass and x-coordinate ( m_i x_i). m_i x_i = (2\,kg × 1) + (3\,kg × 4) + (5\,kg × 2) m_i x_i = 2 + 12 + 10 = 24\,kg·units Step 4: Calculate the x-coordinate of the center of mass (x). x = ( m_i x_i)/( m_i) = 24\,kg·units10\,kg = 2.4\,units Step 5: Calculate the sum of the products of mass and y-coordinate ( m_i y_i). m_i y_i = (2\,kg × 2) + (3\,kg × 1) + (5\,kg × 5) m_i y_i = 4 + 3 + 25 = 32\,kg·units Step 6: Calculate the y-coordinate of the center of mass (y). y = ( m_i y_i)/( m_i) = 32\,kg·units10\,kg = 3.2\,units The center of mass is at (2.4, 3.2). --- Example 2: Composite Lamina (L-shaped) Find the center of mass of a uniform L-shaped lamina formed by two rectangles: • Rectangle 1: Base from x=0 to x=6, height from y=0 to y=2. • Rectangle 2: Base from x=0 to x=2, height from y=2 to y=5. Step 1: Divide the composite shape into simpler geometric shapes and find their individual areas and centroids. • Rectangle 1 (R1): Dimensions: width 6, height 2. Area A_1 = 6 × 2 = 12. Centroid (x_1, y_1): Midpoint of width is 6/2 = 3. Midpoint of height is 2/2 = 1. So, (x_1, y_1) = (3, 1). • Rectangle 2 (R2): Dimensions: width 2, height 3 (from y=2 to y=5). Area A_2 = 2 × 3 = 6. Centroid (x_2, y_2): Midpoint of width is 2/2 = 1. Midpoint of height is 2 + (3/2) = 2 + 1.5 = 3.5. So, (x_2, y_2) = (1, 3.5). Step 2: Calculate the total area of the lamina. A_i = A_1 + A_2 = 12 + 6 = 18 Step 3: Calculate the sum of the products of area and x-coordinate ( A_i x_i). A_i x_i = (A_1 × x_1) + (A_2 × x_2) A_i x_i = (12 × 3) + (6 × 1) = 36 + 6 = 42 Step 4: Calculate the x-coordinate of the center of mass (x). x = ( A_i x_i)/( A_i) = (42)/(18) = (7)/(3) ≈ 2.33 Step 5: Calculate the sum of the products of area and y-coordinate ( A_i y_i). A_i y_i = (A_1 × y_1) + (A_2 × y_2) A_i y_i = (12 × 1) + (6 × 3.5) = 12 + 21 = 33 Step 6: Calculate the y-coordinate of the center of mass (y). y = ( A_i y_i)/( A_i) = (33)/(18) = (11)/(6) ≈ 1.83 The center of mass is at ((7)/(3), (11)/(6)). --- Example 3: Lamina with a Hole Find the center of mass of a square lamina with side length 4\,m (from x=0 to x=4, y=0 to y=4) with a circular hole of radius 1\,m centered at (1, 1). Step 1: Treat the shape as a large square with a "negative mass" (or negative area) circular hole. • Square (S): Dimensions: side length 4\,m. Area A_S = 4 × 4 = 16\,m^2. Centroid (x_S, y_S): Midpoint of width is 4/2 = 2. Midpoint of height is 4/2 = 2. So, (x_S, y_S) = (2, 2). • Circular Hole (H): Radius r = 1\,m. Area A_H = r^2 = (1)^2 = \,m^2. Centroid (x_H, y_H): Given as (1, 1). For calculation, we use this area as negative: -A_H = -. Step 2: Calculate the total effective area. A_i = A_S - A_H = 16 - Step 3: Calculate the sum of the products of area and x-coordinate ( A_i x_i). A_i x_i = (A_S × x_S) + (-A_H × x_H) A_i x_i = (16 × 2) + (- × 1) = 32 - Step 4: Calculate the x-coordinate of the center of mass (x). x = ( A_i x_i)/( A_i) = (32 - )/(16 - ) x ≈ (32 - 3.14159)/(16 - 3.14159) = (28.85841)/(12.85841) ≈ 2.244 Step 5: Calculate the sum of the products of area and y-coordinate ( A_i y_i). A_i y_i = (A_S × y_S) + (-A_H × y_H) A_i y_i = (16 × 2) + (- × 1) = 32 - Step 6: Calculate the y-coordinate of the center of mass (y). y = ( A_i y_i)/( A_i) = (32 - )/(16 - ) y ≈ 2.244 The center of mass is at ((32 - )/(16 - ), (32 - )/(16 - )). What's next? Send 'em! 📸