The centre of mass of a thin rectangular plate (fig - x) with sides of length a and b, whose mass per unit area (sigma)…

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Q4 jee_main_2025_24_jan_evening Rolling Motion

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:

A. $(2)/(5)$ $\frac{2}{5}$
B. $(5)/(2)$ $\frac{5}{2}$
C. $(3)/(4)$ $\frac{3}{4}$
D. $(4)/(3)$ $\frac{4}{3}$

Solution

Related Formula

Klinear = (1)/(2) m vcm²

Krotational = (1)/(2) I ω²

Core Logic

For a solid sphere, the moment of inertia about the center of mass is $I = (2)/(5)mR²$ $I = \frac{2}{5}mR^2$. Since it rolls without slipping, the condition $vcm = ω R$ $v_{\text{cm}} = \omega R$ holds.

Substituting $I$ $I$ and $ω$ $\omega$ into the ratio:

KlinearKrotational = (1)/(2) m vcm²(1)/(2) ((2)/(5)mR²) ( vcmR)²

KlinearKrotational = (1)/((2)/(5)) = (5)/(2)

Pattern Recognition

The ratio of translational to rotational kinetic energy for any rolling body is given by $mR²Icm$ $\frac{mR^2}{I_{\text{cm}}}$. For a solid sphere, this becomes $(1)/(2/5) = (5)/(2)$ $\frac{1}{2/5} = \frac{5}{2}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q9 jee_main_2025_24_jan_evening Rolling on an Inclined Plane

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be

t₁

and

t₂

, respectively, then

A. $t₁ < t₂$ $t_1 < t_2$
B. $t₁ = t₂$ $t_1 = t_2$
C. $t₁ = 2t₂$ $t_1 = 2t_2$
D. $t₁ > t₂$ $t_1 > t_2$

Solution

Related Formula

t = 2 acm

acm = g θ1 + IcmMR²

Core Logic

For a solid sphere: $Isolid = (2)/(5)MR² a₁ = (g θ)/(1 + 2/5) = (5)/(7)g θ$ .

For a hollow sphere: $Ihollow = (2)/(3)MR² a₂ = (g θ)/(1 + 2/3) = (3)/(5)g θ$ .

Comparing accelerations: $a₁ > a₂$ $a_1 > a_2$

Since acceleration of the solid sphere is greater, it takes less time to descend the incline: $t₁ < t₂$ $t_1 < t_2$

Sphere rolling down an incline schematic Q9

Pattern Recognition

Smaller moment of inertia mass distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q jee_main_2025_24_jan_morning Angular Momentum

An object of mass 'm' is projected from origin in a vertical xy plane at an angle

45°

with the x-axis with an initial velocity

v₀

The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]

A. $mv₀³2\sqrt(2)g$ $\frac{mv_{0}^{3}}{2\sqrt{2}g}$ along negative z-axis
B. $mv₀³2\sqrt(2)g$ $\frac{mv_{0}^{3}}{2\sqrt{2}g}$ along positive z-axis
C. $mv₀³4\sqrt(2)g$ $\frac{mv_{0}^{3}}{4\sqrt{2}g}$ along positive z-axis
D. $mv₀³4\sqrt(2)g$ $\frac{mv_{0}^{3}}{4\sqrt{2}g}$ along negative z-axis

Solution

Related Formula

The definition of angular momentum vector $L$ $\vec{L}$ relative to the origin is:

L = r \times p = m( r \times v)

In scalar form for a horizontal speed component at a maximum altitude $H$ $H$:

$L = m vₓ H$ $L = m v_{x} H$

Core Logic

At maximum height, the vertical speed component drops to zero, and the projectile travels entirely horizontally along the curve profile as shown in

Angular Momentum diagram for Q8 - JEE Main 2025 Morning

:

vₓ = v₀ 45° = v₀\sqrt(2)

H = v₀² ² 45°2g = v₀²4g

Step 1: Calculating Magnitude and Vector Direction

Evaluate the horizontal vector cross components:

L = m ( v₀\sqrt(2)) ( v₀²4g) = mv₀³4\sqrt(2)g

Using the right-hand rule, $r$ $\vec{r}$ points into quadrant-1 while velocity points towards $+ i$ $+\hat{i}$. Therefore, $r \times v$ $\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative $k$ $\hat{k}$ orientation (along the negative z-axis).

Pattern Recognition

At peak height, always map $L$ $\vec{L}$ using $m \cdot vpeak \cdot ymax$ . This scalar form simplifies the calculation significantly.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q11 jee_main_2025_24_jan_morning Rolling Motion

A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination

45°

If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-

A. $1\sqrt(2) g$ $\frac{1}{\sqrt{2}} g$
B. $13\sqrt(2) g$ $\frac{1}{3\sqrt{2}} g$
C. $\sqrt(2) g3$ $\frac{\sqrt{2} g}{3}$
D. $\sqrt(2) g$ $\sqrt{2} g$

Solution

Related Formula

The linear acceleration $a$ $a$ for pure rolling motion down an incline profile is:

a = g θ1 + Imr²

Core Logic

For a uniform solid cylinder, the moment of inertia around its central axis is:

I = (1)/(2)mr² Imr² = (1)/(2)

Step 1: Calculating Acceleration

Substitute $θ = 45°$ $\theta = 45^{\circ}$ and the cylinder inertial factor into the formula :

a = g 45°1 + (1)/(2) = g\sqrt(2)(3)/(2)

a = 2g3\sqrt(2) = \sqrt(2)g3

Pattern Recognition

Solid cylinder rolls with acceleration matching $(2)/(3) g θ$ $\frac{2}{3} g\sin\theta$. Since $45° = 1\sqrt(2)$ $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to $\sqrt(2)g3$ $\frac{\sqrt{2}g}{3}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

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

Solution & Explanation

Core Logic

Step 1: Final Position Coordinates

Pattern Recognition

Chapter Mix

Reference Study Guides

More System of Particles and Rotational Motion Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Magnitude and Vector Direction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Acceleration

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Visual Context

Pattern Recognition

Chapter Mix

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