The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R

Q17 jee_main_2025_28_jan_evening Torque and Equilibrium

A uniform rod of mass

250g

having length

100cm

is balanced on a sharp edge at

40cm

mark[cite: 150, 151]. A mass of

400g

is suspended at

10cm

mark. To maintain the balance of the rod, the mass to be suspended at

90cm

mark, is [cite: 154, 156]

A. $300g$ $300\mathrm{g}$
B. $190g$ $190\mathrm{g}$
C. $200g$ $200\mathrm{g}$
D. $290g$ $290\mathrm{g}$

Solution

Related Formula

For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques:

Σ τpivot = 0 Σ (mᵢ \cdot g \cdot xᵢ) = 0

Core Logic

The rod is uniform, meaning its mass ( $250 g$ $250\text{ g}$) acts exactly at its geometric center of mass, the $50 cm$ $50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the $40 cm$ $40\text{ cm}$ mark .

Calculate the relative lever arms from the pivot [cite: 775, 776, 777]:

$400 g$ $400\text{ g}$ mass at $10 cm$ $10\text{ cm}$ mark: lever arm $= 40 - 10 = 30 cm$ $= 40 - 10 = 30\text{ cm}$ (counter-clockwise)
$250 g$ $250\text{ g}$ rod mass at $50 cm$ $50\text{ cm}$ mark: lever arm $= 50 - 40 = 10 cm$ $= 50 - 40 = 10\text{ cm}$ (clockwise)
Unknown mass $M$ $M$ at $90 cm$ $90\text{ cm}$ mark: lever arm $= 90 - 40 = 50 cm$ $= 90 - 40 = 50\text{ cm}$ (clockwise)

Setting up the torque balance equation:

400 \times 30 = (250 \times 10) + (M \times 50)

12000 = 2500 + 50M

50M = 9500 M = (9500)/(50) = 190 g

Step 1: Visual Context

The structural layout of forces acting on the balanced rod system is shown below:

Torque and Equilibrium balancing diagram for Q17

Pattern Recognition

Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q jee_main_2025_29_jan_morning Torque

The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of

F = i - j + k

acts on the particle, then the magnitude of torque (with respect to origin) in

z

-direction is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

τ = r \times F

Core Logic

Given position vector $r = i + j + k$ $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force $F = i - j + k$ $\vec{F} = \hat{i} - \hat{j} + \hat{k}$:

τ = | arrayccc i & j & k 1 & 1 & 1 1 & -1 & 1 array |

Step 1: Isolate z-component

τz = k(1(-1) - 1(1)) = -2 k

The absolute magnitude of the torque component in the $z$ $z$-direction equals $2 ~N \cdot m$ $2 \mathrm{~N \cdot m}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q jee_main_2024_01_february_morning Centre of Mass

The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to

4~m

each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is

4\sqrt(2)x

, where the value of

x

is ______.

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Position vector of the Centre of Mass (COM):

rCOM = m₁ r₁ + m₂ r₂ + m₃ r₃m₁ + m₂ + m₃

Core Logic

Assign coordinates to the three masses ( $m₁=m₂=m₃=2M$ $m_1=m_2=m_3=2M$):

Origin mass: $r₁ = 0 i + 0 j$ $\vec{r}_1 = 0\hat{i} + 0\hat{j}$
X-axis mass: $r₂ = 4 i + 0 j$ $\vec{r}_2 = 4\hat{i} + 0\hat{j}$
Y-axis mass: $r₃ = 0 i + 4 j$ $\vec{r}_3 = 0\hat{i} + 4\hat{j}$

Substitute these into the COM formula:

rCOM = 2M(0) + 2M(4 i) + 2M(4 j)2M + 2M + 2M = 8M i + 8M j6M = (4)/(3) i + (4)/(3) j

Step 1: Calculate Position Vector Magnitude

| rCOM| = \sqrt(((4)/(3))² + ((4)/(3))²) = 4\sqrt(2)3

Matching this directly with the given template $4\sqrt(2)x$ $\frac{4\sqrt{2}}{x}$ shows that $x = 3$ $x = 3$.

Pattern Recognition

Since the mass layout is completely symmetric along both right-angle legs, the COM coordinates are identical ( $xCOM = yCOM$ $x_{\text{COM}} = y_{\text{COM}}$).

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q60 jee_main_2024_29_january_evening Angular Momentum of a Particle

A body of mass

5 kg

moving with a uniform speed

3\sqrt(2) ms⁻¹

in X–Y plane along the line

y = x + 4

. The angular momentum of the particle about the origin will be ______

kg m²s⁻¹

.

Numerical Answer. Answer: 60 to 60

Solution

Related Formula

The magnitude of the angular momentum $L$ $L$ of a particle of mass $m$ $m$ moving with velocity $v$ $v$ is:

$L = m v d$ $L = m v d$

where:

$d$ $d$ is the perpendicular distance from the axis of rotation (origin) to the line of motion of the particle.

Core Logic

Given parameters:

Mass, $m = 5 kg$ $m = 5\text{ kg}$
Velocity, $v = 3\sqrt(2) ms⁻¹$ $v = 3\sqrt{2}\text{ ms}^{-1}$
Line of motion: $y = x + 4 x - y + 4 = 0$ $y = x + 4 \implies x - y + 4 = 0$

Step 1: Calculate Perpendicular Distance

The perpendicular distance $d$ $d$ from the origin $(0,0)$ $(0,0)$ to the line $Ax + By + C = 0$ $Ax + By + C = 0$ is:

d = |A(0) + B(0) + C|\sqrt(A² + B²)

For the line $x - y + 4 = 0$ $x - y + 4 = 0$:

d = |4|\sqrt(1² + (-1)²) = 4\sqrt(2) = 2\sqrt(2) m

Step 2: Calculate Angular Momentum

Substitute the values into the angular momentum formula:

$L = m v d$ $L = m v d$

L = 5 kg \times (3\sqrt(2) ms⁻¹) \times (2\sqrt(2) m)

L = 5 \times 3 \times 4 = 60 kg m²s⁻¹

Thus, the angular momentum of the particle about the origin is $60 kg m²s⁻¹$ $60\text{ kg m}^2\text{s}^{-1}$.

Pattern Recognition

Instead of complicated vector cross products, find the perpendicular distance of the straight line from the origin using standard coordinate geometry. $L = mvd$ $L = mvd$ is extremely fast and reliable.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q jee_main_2024_27_jan_morning Moment of Inertia

Four particles each of mass

1 kg

are placed at four corners of a square of side

2 m

. The moment of inertia of the system about an axis perpendicular to its plane and passing through one of its vertices is ______

kg ²

. {{IMG}}

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

I = Σ mᵢ rᵢ²

Core Logic

Let the axis pass through vertex 1. Evaluate distances ( $r$ $r$) for each corner particle:

Particle at vertex 1: $r₁ = 0$ $r_1 = 0$
Particle at adjacent vertex 2: $r₂ = a$ $r_2 = a$
Particle at adjacent vertex 4: $r₄ = a$ $r_4 = a$
Particle at diagonally opposite vertex 3: $r₃ = \sqrt(2)a$ $r_3 = \sqrt{2}a$

Step 1: Set up substitution formula

I = m(0)² + m(a)² + m(a)² + m(\sqrt(2)a)²

I = ma² + ma² + 2ma² = 4ma²

Step 2: Numeric Evaluation

Substitute $m = 1 kg$ $m = 1\text{ kg}$ and side length $a = 2 m$ $a = 2\text{ m}$:

I = 4 \times 1 \times (2)² = 4 \times 4 = 16 kg ²

Pattern Recognition

For a standard planar configuration system, total orthogonal moment components map predictably via basic summation configurations matching $4ma²$ $4ma^2$ exactly.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R < L) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M):

Solution & Explanation

Related Formula

Core Logic

Step 1: Moment of Inertia for cylinders bisected perpendicularly

Step 2: Moment of Inertia for cylinders parallel to axis

Step 3: Total Moment of Inertia

Step 4: Substitute Total Mass

Pattern Recognition

Chapter Mix

Reference Study Guides

More Systems of Particles and Rotational Motion Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Visual Context

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Isolate z-component

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Position Vector Magnitude

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Perpendicular Distance

Step 2: Calculate Angular Momentum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up substitution formula

Step 2: Numeric Evaluation

Pattern Recognition

Chapter Mix

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