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An object as shown in the figure consists of a sphere of mass $1 \mathrm{~kg}$ and radius $5 \mathrm{~cm}$ attached to a rod of length $50 \mathrm{~cm}$ and mass $500 \mathrm{~g}$. If this has to be balanced horizontally on a knife-edge, at what distance from point A must the knife-edge be placed?

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Verified Answer
The correct answer is:
45 cm
CONCEPT:
- Center of mass: Center of the mass of a body is the weighted average position of all the parts of the body with respect to mass.
- The Center of mass is used in representing irregular objects as point masses for the ease of calculation.
- For simple shaped objects, its center of mass lies at the centroid.
- For irregular shapes, the center of mass is found by the vector addition of the weighted position vectors.
The position coordinates for the center of mass can be found by:
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n} \quad C_y=\frac{m_1 y_1+m_2 y_2+\ldots m_n y_n}{m_1+m_2+\ldots m_n}$
CALCULATION:
The center of mass of an object is the point at which the object can be balanced. Hence, the given object can be balanced by placing a knife edge under the center of mass.
Given that:
Mass of the rod, $m_A=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Mass of the sphere, $m_B=1 \mathrm{~kg}$
$\begin{aligned} & x_1=25 \mathrm{~cm} \\ & x_2=50+5=55 \mathrm{~cm}\end{aligned}$

To find the coordinates of the centre of mass of the object from $A$,
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n}=\frac{m_A x_A+m_B x_B}{m_A+m_B}=\frac{(0.5 \times 25)+(1 \times 55)}{0.5+1}=45 \mathrm{~cm}$
Hence, the knife-edge must be placed $45 \mathrm{~cm}$ from point $A$.
- Center of mass: Center of the mass of a body is the weighted average position of all the parts of the body with respect to mass.
- The Center of mass is used in representing irregular objects as point masses for the ease of calculation.
- For simple shaped objects, its center of mass lies at the centroid.
- For irregular shapes, the center of mass is found by the vector addition of the weighted position vectors.
The position coordinates for the center of mass can be found by:
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n} \quad C_y=\frac{m_1 y_1+m_2 y_2+\ldots m_n y_n}{m_1+m_2+\ldots m_n}$
CALCULATION:
The center of mass of an object is the point at which the object can be balanced. Hence, the given object can be balanced by placing a knife edge under the center of mass.
Given that:
Mass of the rod, $m_A=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Mass of the sphere, $m_B=1 \mathrm{~kg}$
$\begin{aligned} & x_1=25 \mathrm{~cm} \\ & x_2=50+5=55 \mathrm{~cm}\end{aligned}$

To find the coordinates of the centre of mass of the object from $A$,
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n}=\frac{m_A x_A+m_B x_B}{m_A+m_B}=\frac{(0.5 \times 25)+(1 \times 55)}{0.5+1}=45 \mathrm{~cm}$
Hence, the knife-edge must be placed $45 \mathrm{~cm}$ from point $A$.
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