How long does a slipping cylinder take to start rolling with v₀=49m/s, ω₀=v₀/4R, μₖ=0.25?

5 seconds. Using v(t)=v₀−μₖgt and ω(t)R=v₀/4+2μₖgt, setting equal: 36.75=7.35t, t=5s.

What is the rolling condition for a cylinder?

v=ωR, where v is translational velocity and ω is angular velocity.

Solid cylinder v₀=49m/s translational, ω₀=v₀/4R rotational, μₖ=0.25 — time to start rolling

JEERANKERS

PhysicsRotation

QA solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At $t=0$: translational velocity $v_0=49$ m/s and rotational velocity $\omega_0=v_0/4R$ about centre. Time taken to start rolling? ($\mu_K=0.25$, $g=9.8$ m/s²)

A) 15 s B) 5 s C) 10 s D) 7.5 sRotational Motion

MCQ

✅ Correct Answer

B) 5 s

✓

Solution

Equations for translation and rotation

Friction $f=\mu_K mg$ acts backward (opposing motion), providing torque to spin up.

Translation: $v(t)=v_0-\mu_K g\cdot t=49-0.25\times9.8\times t=49-2.45t$

Rotation: $I\alpha=fR\Rightarrow\dfrac{mR^2}{2}\cdot\alpha=\mu_K mg\cdot R$
$\alpha=\dfrac{2\mu_K g}{R}$

$\omega(t)=\omega_0+\alpha t=\dfrac{v_0}{4R}+\dfrac{2\mu_K g}{R}\cdot t$

Rolling condition: v = ωR

$49-2.45t=\left(\dfrac{v_0}{4R}+\dfrac{2\mu_K g}{R}\cdot t\right)\cdot R$

$49-2.45t=\dfrac{v_0}{4}+2\mu_K g\cdot t$

$49-2.45t=\dfrac{49}{4}+4.9t$

$49-12.25=4.9t+2.45t=7.35t$

$36.75=7.35t\Rightarrow t=\boxed{5\text{ s}}$

📘 Rolling Without Slipping Condition

For rolling to commence: translational velocity $v=\omega R$. Linear deceleration $=\mu_K g$. Angular acceleration $\alpha=2\mu_K g/R$ (for solid cylinder $I=mR^2/2$). Set $v_0-\mu_K g\cdot t=\omega_0 R+2\mu_K g\cdot t$ and solve for $t$.

💡

Important Concepts

Solid Cylinder Moment of InertiaFor solid cylinder: I=mR²/2. Angular acceleration from friction torque: α=fR/I=μₖmgR/(mR²/2)=2μₖg/R.

Translation DecelerationFriction f=μₖmg decelerates translation: a=μₖg. v(t)=v₀−μₖgt=49−2.45t.

Angular AccelerationFriction also spins up the cylinder: τ=fR, I·α=μₖmgR, α=2μₖg/R. ω(t)=ω₀+αt=v₀/4R+2μₖgt/R.

Rolling Conditionv=ωR: 49−2.45t=(v₀/4R+2μₖg·t/R)·R=v₀/4+2μₖg·t=12.25+4.9t. Solving: 36.75=7.35t, t=5s.

FAQs

Why does friction decelerate translation but accelerate rotation?

⌄

Friction acts backward (opposing forward translation), decelerating the cylinder. The same friction creates a torque in the direction of spin (forward rolling), spinning the cylinder up.

What is ω₀ in terms of numbers?

⌄

ω₀=v₀/4R=49/4R. When multiplied by R: ω₀R=49/4=12.25 m/s. This is the initial peripheral speed.

What speed does the cylinder reach when rolling?

⌄

v=49−2.45×5=49−12.25=36.75 m/s. Check: ωR=(v₀/4+2μₖg×5)=(12.25+4.9×5)R/R=36.75 m/s ✓

Would a hollow cylinder take longer?

⌄

Yes. Hollow cylinder I=mR², so α=μₖg/R (half the angular acceleration). It takes longer to achieve rolling.

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026.

🔗