Worked examples · Sheet 2

Centers of mass and motion of the center of mass

Theory used: Center of mass Center of mass of a continuous body Center of mass of a composite body Center-of-mass theorem System of particles

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Three particles have masses $m_{1} = 1.0 kg$ , $m_{2} = 2.0 kg$ , $m_{3} = 3.0 kg$ and are located at the points of the plane $P_{1} = (0, 0)$ , $P_{2} = (3.0, 0)$ , $P_{3} = (1.0, 2.0)$ , with coordinates in meters. Determine the position of the center of mass.

x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{m _{1} + m _{2} + m _{3}}, y_{c m} = \frac{m _{1} y _{1} + m _{2} y _{2} + m _{3} y _{3}}{m _{1} + m _{2} + m _{3}}

M = 1.0 + 2.0 + 3.0 = 6.0 kg

x_{c m} = \frac{1.0 \cdot 0 + 2.0 \cdot 3.0 + 3.0 \cdot 1.0}{6.0} m = \frac{9.0}{6.0} m = 1.5 m

y_{c m} = \frac{1.0 \cdot 0 + 2.0 \cdot 0 + 3.0 \cdot 2.0}{6.0} m = \frac{6.0}{6.0} m = 1.0 m

Each coordinate of the center of mass is the average of the coordinates of the particles weighted by the masses, so $x_{c m}$ and $y_{c m}$ are computed separately by dividing by the total mass $6.0 kg$ . The points $P_{1}$ and $P_{2}$ lie on the $x$ axis and do not contribute to the numerator of $y_{c m}$ . The center of mass is therefore $(1.5 m, 1.0 m)$ .

As a check, the point found is inside the triangle $P_{1} P_{2} P_{3}$ , as it must be for a system of positive masses, and it is closer to $P_{3}$ , which carries the largest mass, than to $P_{1}$ and $P_{2}$ , from which it is equidistant.

Exercise 2

From a uniform square sheet of side $L = 4.0 m$ a square of side $\frac{L}{2}$ is cut out at one vertex. Determine the center of mass of the remaining sheet, in a reference frame with origin at the vertex opposite to the cut one and axes along two sides.

The remaining sheet (shaded) is the whole square with center

C_{0}

minus the dashed square with center

C_{1}

; the center of mass has

x_{c m} = y_{c m} = \frac{5}{3} m

and lies on the diagonal, on the side of

C_{0}

opposite to

C_{1}

σ = \frac{M _{0}}{L ^{2}}, M_{1} = σ \frac{L ^{2}}{4} = \frac{M _{0}}{4}, x_{c m} = \frac{M _{0} x _{0} - M _{1} x _{1}}{M _{0} - M _{1}}

x_{c m} = \frac{M _{0} \cdot \frac{L}{2} - \frac{M _{0}}{4} \cdot \frac{3 L}{4}}{\frac{3 M _{0}}{4}} = \frac{\frac{L}{2} - \frac{3 L}{16}}{\frac{3}{4}} = \frac{5 L}{16} \cdot \frac{4}{3} = \frac{5 L}{12}

x_{c m} = y_{c m} = \frac{5 \cdot 4.0 m}{12} \approx 1.7 m

The sheet is uniform, so the mass of each part is proportional to its area and the cut-out square has one quarter of the mass $M_{0}$ of the whole square. By the theorem on the center of mass of a composite body, the whole square consists of the remaining sheet and the cut-out square; therefore $M_{0} x_{0} = (M_{0} - M_{1}) x_{c m} + M_{1} x_{1}$ , and from this relation we obtain the formula with the mass of the missing piece subtracted. The centers of mass of the two squares are their centers, by symmetry, so $x_{0} = \frac{L}{2} = 2.0 m$ and $x_{1} = \frac{3 L}{4} = 3.0 m$ .

The figure is symmetric with respect to the diagonal through the origin, so $y_{c m} = x_{c m}$ . The center of mass is at $(\frac{5}{3} m, \frac{5}{3} m)$ , that is approximately at $(1.7 m, 1.7 m)$ .

As a check, the remaining sheet can be divided into two rectangles, one $4.0 m \times 2.0 m$ with center $(2.0, 1.0)$ and one $2.0 m \times 2.0 m$ with center $(1.0, 3.0)$ . The areas are $8.0 m^{2}$ and $4.0 m^{2}$ , and we obtain $x_{c m} = \frac{8 \cdot 2 + 4 \cdot 1}{12} m = \frac{5}{3} m$ , the same value.

Exercise 3

A straight rod of length $L = 1.8 m$ lies on the $x$ axis between $x = 0$ and $x = L$ and has linear density

λ (x) = λ_{0} (1 + \frac{x}{L}),

with $λ_{0}$ a positive constant. Compute the mass of the rod as a function of $λ_{0}$ and the position of the center of mass.

M = \int_{0}^{L} λ (x) d x, x_{c m} = \frac{1}{M} \int_{0}^{L} x λ (x) d x

M = λ_{0} \int_{0}^{L} (1 + \frac{x}{L}) d x = λ_{0} [x + \frac{x ^{2}}{2 L}]_{0}^{L} = λ_{0} (L + \frac{L}{2}) = \frac{3}{2} λ_{0} L

\int_{0}^{L} x λ (x) d x = λ_{0} \int_{0}^{L} (x + \frac{x ^{2}}{L}) d x = λ_{0} [\frac{x ^{2}}{2} + \frac{x ^{3}}{3 L}]_{0}^{L} = \frac{5}{6} λ_{0} L^{2}

x_{c m} = \frac{\frac{5}{6} λ _{0} L ^{2}}{\frac{3}{2} λ _{0} L} = \frac{5}{9} L = \frac{5}{9} \cdot 1.8 m = 1.0 m

The rod is not uniform, so its center of mass does not coincide with the midpoint and must be computed with the definition for a continuous body. A segment of length $d x$ at position $x$ has mass $d m = λ (x) d x$ ; the total mass is the integral of $λ$ over the rod, and the center of mass is the ratio between the integral of $x λ (x)$ and the mass.

The two integrals are integrals of polynomials and are computed with antiderivatives. The factor $λ_{0}$ appears both in the numerator and in the denominator, so it cancels and the center of mass depends only on $L$ ; it is located at $x_{c m} = \frac{5}{9} L = 1.0 m$ .

As a check, the density increases from $λ_{0}$ at $x = 0$ to $2 λ_{0}$ at $x = L$ , therefore the center of mass must lie beyond the midpoint $0.9 m$ , on the denser side; the result $1.0 m$ satisfies this condition.

Exercise 4

Two blocks of masses $m_{1} = 2.0 kg$ and $m_{2} = 3.0 kg$ , at rest on a smooth horizontal plane, are connected by a spring of negligible mass. From $t = 0$ a constant horizontal force of magnitude $F = 10 N$ , directed along the spring, is applied to block $1$ . Compute the acceleration of the center of mass, the displacement of the center of mass after $3.0 s$ and the acceleration of block $2$ at the instant when block $1$ has acceleration $a_{1} = 3.5 m/ s^{2}$ in the direction of $F$ .

(m_{1} + m_{2}) a_{c m} = F ⟹ a_{c m} = \frac{F}{m _{1} + m _{2}} = \frac{10 N}{5.0 kg} = 2.0 m/ s^{2}

Δ x_{c m} = \frac{1}{2} a_{c m} t^{2} = \frac{1}{2} \cdot 2.0 m/ s^{2} \cdot (3.0 s)^{2} = 9.0 m

(m_{1} + m_{2}) a_{c m} = m_{1} a_{1} + m_{2} a_{2} ⟹ a_{2} = \frac{F - m _{1} a _{1}}{m _{2}}

a_{2} = \frac{10 N - 2.0 kg \cdot 3.5 m/ s ^{2}}{3.0 kg} = \frac{3.0 N}{3.0 kg} = 1.0 m/ s^{2}

We choose as the system the two blocks together with the spring. The elastic force of the spring is internal, while the external forces are the weights, the reactions of the plane and the force $F$ ; the plane is horizontal and smooth, so weights and reactions are vertical and balance each other, and the resultant of the external forces is $F$ . By the center-of-mass theorem the center of mass has constant acceleration $\frac{F}{M}$ , like a single particle of mass $5.0 kg$ pushed by the same force, whatever the extension of the spring.

The system starts from rest, so the center of mass moves with uniformly accelerated motion with zero initial velocity and in $3.0 s$ it is displaced by $9.0 m$ . The individual blocks instead oscillate relative to each other, and their accelerations change in time.

At every instant $M a_{c m} = m_{1} a_{1} + m_{2} a_{2}$ , because this is the second derivative of the definition of the center of mass. Given $a_{1}$ , the equality yields $a_{2} = 1.0 m/ s^{2}$ in the direction of $F$ . At that instant the spring exerts on block $2$ a force of $3.0 N$ in the direction of $F$ , because $m_{2} a_{2} = 3.0 N$ .