A cylinder is rolling down on an inclined plane of inclination 60. It's acceleration during rolling down will be…

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Q12 jee_main_2025_04_april_evening Rolling Motion

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be:

A. $4\sqrt(2) m/s$ $4\sqrt{2}\text{ m/s}$
B. $8 m/s$ $8\text{ m/s}$
C. $4 m/s$ $4\text{ m/s}$
D. $8\sqrt(2) m/s$ $8\sqrt{2}\text{ m/s}$

Solution

Related Formula

Velocity of a point on a rolling wheel at angular position $θ$ $\theta$ from the lowest point:

v = 2 vcm ((θ)/(2))

Core Logic

At the highest point, $θ = 180^°$ $\theta = 180^\circ$, so:

vₜₒₚ = 2vcm = 8 m/s vcm = 4 m/s

For a particle on the rim at the same horizontal level as the center, the angle from the lowest point is $θ = 90^°$ $\theta = 90^\circ$.

Step 1: Compute Speed at Mid-Height

Substituting $θ = 90^°$ $\theta = 90^\circ$ into our velocity relation: $vmid = 2vcm (45^°) = 2 \times 4 \times 1\sqrt(2) = 4\sqrt(2) m/s$

Instantaneous center of rotation on rolling wheel

Pattern Recognition

The contact point with the ground is the Instantaneous Center of Rotation (ICR). Distance to top point is $2R$ $2R$, distance to mid-level point is $\sqrt(R²+R²) = \sqrt(2)R$ $\sqrt{R^2+R^2} = \sqrt{2}R$. Velocity scales linearly with distance from ICR.

Chapter Mix

Class 11 Physics: Rotational Motion

Q22 jee_main_2025_04_april_evening Conservation of Angular Momentum

A solid sphere with uniform density and radius R is rotating initially with constant angular velocity

(ω₁)

about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius becomes

R / 2

is

xω₁

. The value of x is ________.

Numerical Answer. Answer: 32 to 32

Solution

Related Formula

Conservation of Angular Momentum (since no external torque acts):

I₁ ω₁ = I₂ ω₂

For a solid sphere, moment of inertia is:

I = (2)/(5)MR²

Mass scales with volume: $M \propto R³$ $M \propto R^3$

Core Logic

When the radius reduces to $R₂ = (R)/(2)$ $R_2 = \frac{R}{2}$, the mass scales cubically:

M₂ = M₁ ((R/2)/(R))³ = (M₁)/(8)

Now, compute the new moment of inertia $I₂$ $I_2$:

I₂ = (2)/(5) M₂ R₂² = (2)/(5) ((M₁)/(8)) ((R)/(2))² = (2)/(5) M₁ R² \times (1)/(32) = (I₁)/(32)

Step 1: Compute Final Angular Velocity

Using conservation of angular momentum:

I₁ ω₁ = ((I₁)/(32)) ω₂ ω₂ = 32 ω₁

Hence, the value of $x$ $x$ is 32.

Pattern Recognition

Since inertia of a solid sphere scales with $M R²$ $M R^2$ and $M \propto R³$ $M \propto R^3$, the net moment of inertia scales with $R⁵$ $R^5$. Shrinking the radius by half ( $1/2$ $1/2$) cuts down inertia by a factor of $(1/2)⁵ = 1/32$ $(1/2)^5 = 1/32$. Velocity must scale up by 32 to conserve momentum.

Chapter Mix

Class 11 Physics: Rotational Motion

Q jee_main_2025_04_april_morning Torque and Angular Momentum

Which of the following are correct expressions for torque acting on a body? A.

τ = r \times L

B.

τ = (d)/(dt)( r \times p)

C.

τ = r \times d pdt

D.

τ = I α

E.

τ = r \times F

( r = position vector; p = linear momentum; L = angular momentum; α = angular acceleration; I = moment of inertia; F = force; t = time)

Choose the correct answer from the options given below:

A. B, D and E Only
B. C and D Only
C. B, C, D and E Only
D. A, B, D and E Only

Solution

Related Formula

Fundamental mathematical definition of torque:

τ = r \times F

Rotational analogue of Newton's second law:

τ = d Ldt = I α

Linear momentum relations:

L = r \times p τ = (d)/(dt)( r \times p)

Core Logic

Let's check each expression sequentially:

A. $τ = I \times L$ $\vec{\tau} = \vec{I} \times \vec{L}$ is dimensionally incorrect (Moment of inertia $I$ $I$ is primarily treated as a tensor or scalar placeholder, not crossed directly like this).
B. $τ = d Ldt = (d)/(dt)( r \times p)$ is fundamentally correct.
C. $τ = r \times F = r \times d pdt$ is correct since $F = d pdt$ $\vec{F} = \frac{d\vec{p}}{dt}$.
D. $τ = I α$ $\vec{\tau} = I\vec{\alpha}$ is the standard scalar component/fixed axis formulation.
E. $τ = r \times F$ $\vec{\tau} = \vec{r} \times \vec{F}$ is the true physical vector definition.

Thus, statements B, C, D, and E are universally correct representations.

Pattern Recognition

Torque can be represented either through geometric structural parameters (position and force cross products) or via kinematic response properties (rate of change of angular momentum or product of rotational inertia and acceleration).

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q jee_main_2025_04_april_morning Uniform Circular Motion and Dynamics

If

L

and

P

represent the angular momentum and linear momentum respectively of a particle of mass '

m

' having position vector

r = a( i ω t + j ω t)

. The direction of force is

A. Opposite to the direction of $r$ $\vec{r}$
B. Opposite to the direction of $L$ $\vec{L}$
C. Opposite to the direction of $P$ $\vec{P}$
D. Opposite to the direction of $L \times P$ $\vec{L} \times \vec{P}$

Solution

Related Formula

Acceleration vector equation via secondary derivation:

a = d² rdt²

Force equation:

F = m a

Core Logic

Given position tracking trace:

r = a( i ω t + j ω t)

Velocity vector $v$ $\vec{v}$:

v = d rdt = aω(- i ω t + j ω t)

Step 1: Differentiate to find Acceleration

a = d vdt = aω²(- i ω t - j ω t)

a = -ω² [ a( i ω t + j ω t) ] = -ω² r

Step 2: Establish Force Direction

F = m a = -mω² r

The minus sign indicates the net centripetal pulling force aligns explicitly opposite to the direction of $r$ $\vec{r}$.

Pattern Recognition

The expression describes a standard uniform circular motion profile. In circular configurations, acceleration and centripetal forces point radially inward, directly opposing the outbound position tracker vector.

Chapter Mix

Class 11 Physics: Kinematics Class 11 Physics: System of Particles and Rotational Motion

Q21 jee_main_2025_04_april_morning Rolling Motion on an Inclined Plane

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is

\sqrt((x)/(5))

where x =

Numerical Answer. Answer: 3.5 to 4

Solution

Related Formula

Velocity of a rolling body from mechanical energy conservation:

v = \sqrt((2gh)/(1 + (k²)/(R²)))

Core Logic

Evaluate radius of gyration factor coefficients:

For a circular ring: $(k²)/(R²) = 1$ $\frac{k^2}{R^2} = 1$
For a solid sphere: $(k²)/(R²) = (2)/(5)$ $\frac{k^2}{R^2} = \frac{2}{5}$

Step 1: Compute Velocity Expressions

vring = \sqrt((2gh)/(1 + 1)) = \sqrt(gh)

vsphere = \sqrt((2gh)/(1 + (2)/(5))) = \sqrt((10gh)/(7))

Step 2: Calculate the Velocity Ratio

vringvsphere = \sqrt(gh)\sqrt((10gh)/(7)) = \sqrt((7)/(10)) = \sqrt((3.5)/(5))

Matching with the prompt expression format $\sqrt((x)/(5))$ $\sqrt{\frac{x}{5}}$ reveals: $x = 3.5$ $x = 3.5$

Rounding to the nearest integer yields 4.

Pattern Recognition

Objects with lower mass concentration near the center (lower $(k²)/(R²)$ $\frac{k^2}{R^2}$ like the sphere) convert gravitational potential energy into translational kinetic energy more efficiently, rolling faster than hollow equivalents.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

A cylinder is rolling down on an inclined plane of inclination 60°$60^{\circ}$. It's acceleration during rolling down will be x√(3) ~m / s²$\frac{x}{\sqrt{3}} \mathrm{~m / s^{2}}$, where x =$x =$ ________ (use g = 10 ~m/s²$g = 10 \mathrm{~m/s}^{2}$).

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Linear Acceleration

Step 2: Solve for x

Step 2: Solve for x

Pattern Recognition

Pattern Recognition

Chapter Mix

More System of Particles and Rotational Motion Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Compute Speed at Mid-Height

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Final Angular Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Differentiate to find Acceleration

Step 2: Establish Force Direction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Velocity Expressions

Step 2: Calculate the Velocity Ratio

Pattern Recognition

Chapter Mix

More System of Particles and Rotational Motion Questions — jee_main_2024_29_jan_morning

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A cylinder is rolling down on an inclined plane of inclination $60°$ $60^{\circ}$. It's acceleration during rolling down will be $x\sqrt(3) ~m / s²$ , where $x =$ $x =$ ________ (use $g = 10 ~m/s²$ $g = 10 \mathrm{~m/s}^{2}$).