A T shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is appl

Q: Why must the force pass through the Center of Mass for pure translation?

Torque is what causes an object to rotate. The formula for torque is $\tau = r \times F$, where $r$ is the distance from the center of mass to the line of action of the force. If the force passes exactly through the center of mass, $r$ is zero, making the torque zero. With zero torque, there is no rotation, only translation.

Q: How did we know the mass of the vertical rod is $2m$?

The problem states it's a 'T shaped object', implying it's made of the same uniform material throughout. For uniform objects, mass is directly proportional to length. Since the vertical rod ($2l$) is twice as long as the horizontal rod ($l$), it must have twice the mass.

Q: Why is the y-coordinate of the vertical rod's center of mass $-l$?

We placed our origin $(0,0)$ at the junction. The vertical rod extends downwards from $y = 0$ to $y = -2l$. The center of mass of a uniform rod is exactly in its middle. The midpoint between $0$ and $-2l$ is $-l$.

Q: Why did we take the absolute value at the end to find PC?

Coordinates can be negative depending on where we place our origin, but 'distance' is a scalar quantity and must always be positive. Taking the absolute difference between the coordinates ensures we get a positive physical length.

Q: What would happen if the force was applied at point C?

If the force was applied at C, it would not pass through the center of mass. This would create a significant torque, causing the T-shape to rotate (spin) while it translates across the floor.

Question

A T shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

Options:

(A) $\frac{2}{3}l$

(B) $\frac{3}{2}l$

(C) $\frac{4}{3}l$

(D) $l$

Sigma Percentile · Accepted Answer

### The Secret to Pure Translation

Imagine sliding a book across a table. If you push it right in the middle, it glides straight forward. But if you push it near the edge, it spins as it moves. This simple observation is the key to solving our problem.

For any rigid body to undergo **pure translational motion** (meaning it moves without spinning or rotating), the net torque acting on it must be exactly zero. The only way a single applied force can produce zero torque is if its line of action passes directly through the **Center of Mass (CM)** of the object.

Therefore, the point $P$ where the force $F$ is applied must be the center of mass of the T-shaped object.

### Setting Up the Geometry

To find the center of mass, we need a coordinate system. Let's place our origin $(0,0)$ right at the junction where the two rods meet.

Because the object is perfectly symmetric along the vertical axis, we know intuitively that the x-coordinate of the center of mass is zero ($x_{cm} = 0$). Our mission is simply to find the y-coordinate using the standard formula:

$$
y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}
$$

### Mass and Coordinates of the Rods

Assuming the object is made of a uniform material, the mass of each section is directly proportional to its length.

**1. The Horizontal Rod:**
It has a length of $l$. Let's assign it a mass of $m_1 = m$. Because it lies exactly on the x-axis, its center of mass is at the origin, so $y_1 = 0$.

**2. The Vertical Rod:**
It has a length of $2l$. Since it's twice as long, its mass will be twice as much, so $m_2 = 2m$. The center of mass of a uniform rod is at its midpoint. Since it extends from $y = 0$ down to $y = -2l$, its midpoint is at $y_2 = -l$.

### Calculating the Center of Mass

Now, we plug these values into our formula:

$$
y_{cm} = \frac{m(0) + 2m(-l)}{m + 2m}
$$

$$
y_{cm} = \frac{-2ml}{3m} = -\frac{2l}{3}
$$

This tells us that the center of mass (point $P$) is located at a distance of $\frac{2l}{3}$ below the junction.

### The Final Trap

Many students find $-\frac{2l}{3}$ and immediately look for it in the options. But we must read the question carefully! It asks for the location of $P$ **with respect to point $C$**, not the junction.

Point $C$ is at the very bottom of the vertical rod, which corresponds to the coordinate $y_C = -2l$. To find the distance $PC$, we calculate the absolute difference between their y-coordinates:

$$
PC = |y_C - y_P|
$$

$$
PC = \left|-2l - \left(-\frac{2l}{3}\right)\right|
$$

$$
PC = \left|-2l + \frac{2l}{3}\right| = \left|-\frac{4l}{3}\right| = \frac{4l}{3}
$$

And there we have it! The force must be applied at a distance of $\frac{4l}{3}$ from the bottom point $C$ to ensure the object glides smoothly without rotating.

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Visualized Solution

$Condition for Pure Translation$

$Coordinate System Setup$

$Mass and CM of Individual Rods$

$Substituting Values$

$Calculating the Center of Mass$

$Finding Distance PC$

$The Way Forward$

The Sigma Insight: Centre of Mass

The Secret to Pure Translation

Setting Up the Geometry

Mass and Coordinates of the Rods

Calculating the Center of Mass

The Final Trap

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Pure Translation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinate System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass and CM of Individual Rods

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Center of Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Distance PC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward