What is the angular displacement made by a particle after 5 s, when it starts from rest with an angular accele

Question 59 of 87

Q. What is the angular displacement made by a particle after 5 s, when it starts from rest with an angular acceleration 0.2 rad s^-2 ?

(a) 4 rad

(b) 1 rad

(d) 5 rad

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2019MCQ· 1mImportance★★★★★

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Concept understanding — Angular Velocity

Angular Velocity: The Language of Spinning

Imagine you're watching a ceiling fan. You know it's moving, but how do you describe how fast it's spinning? You could say "it makes 3 full turns every second" — that's a measure of angular velocity. But let's build this idea from the ground up.

The Intuition: Speed vs. Turning Speed

When a car moves in a straight line, we talk about its linear velocity — how many meters it covers per second. But when something rotates — a wheel, a planet, a spinning top — every point on it moves in a circle. The outer edge of a wheel travels a much longer distance in one rotation than a point near the centre. So if we tried to use ordinary speed (metres per second), we'd get different numbers for different parts of the same object. That's messy.

What we need is a quantity that describes the rotation itself, independent of how far a point is from the centre. That quantity is angular velocity.

The Core Idea

Angular velocity tells you how fast the angle is changing as something rotates. Instead of "metres per second," it's "radians per second" (or degrees per second, or revolutions per second).

Note

A radian is the natural unit for angles in physics. One full circle = $2\pi$ radians ≈ 6.28 rad. So "1 radian per second" means the object sweeps out an angle of about 57.3° every second.

The Precise Definition

Let an object rotate about a fixed axis. At time $t$ , let its angular position be $\theta(t)$ — the angle it has turned through from some reference line. Then:

\omega = \frac{d\theta}{dt}

where $\omega$ (Greek letter omega) is the instantaneous angular velocity. For uniform rotation (constant speed), this simplifies to:

\omega = \frac{\Delta \theta}{\Delta t}

Units: radians per second (rad/s). In practice, you'll also see revolutions per minute (rpm) — 1 rpm = $\frac{2\pi}{60}$ rad/s.

Direction Matters: Angular Velocity as a Vector

Here's where it gets interesting. Angular velocity isn't just a number — it has a direction. But the direction isn't "clockwise" or "anticlockwise" in the plane of rotation. Instead, it points along the axis of rotation, following the right-hand rule:

Tip

Curl the fingers of your right hand in the direction of rotation. Your thumb points in the direction of the angular velocity vector $\vec{\omega}$ .

So a spinning wheel's angular velocity vector points straight out from its axle. If the wheel spins faster, the vector gets longer. If it reverses direction, the vector flips.

Connecting to Linear Velocity

Here's the payoff: once you know the angular velocity of a rotating object, you can find the linear speed of any point on it. For a point at distance $r$ from the axis:

v = \omega r

This is why the outer edge of a merry-go-round moves faster than a point near the centre — same $\omega$ , different $r$ .

Watch out

This formula $v = \omega r$ only works when $v$ is the tangential speed (perpendicular to the radius). It does NOT apply to radial motion (straight in or out).

A Concrete Example …

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Starting from rest, the angular displacement after 5 s under angular acceleration 0.2 rad/s^2 is 2.5 rad.

For rotational motion starting from rest (omega0 = 0) with constant angular acceleration alpha, the angular displacement after time t is given by

theta = omega0*t + (1/2)alphat^2

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Relation between linear velocity and angular velocity of any body is:$
(a) v→ = r→ × ω→
(b) v→ = ω→ × r→
(c) ω→ = v→ × r→
(d) r→ × v→ = ω→
›Reveal solutionSolution

The linear velocity of a rotating particle is v→ = ω→ × r→, obtained from the vector relation between angular and linear velocity.

Consider a rigid body rotating with angular velocity ω→ about a fixed axis. A particle in the body at position vector r→ (measured from a point on the axis) moves in a circle of radius r sinθ (θ = angle between ω→ and r→) with speed v = ωr sinθ. The direction of v→ is perpendicular to both ω→ and r→, given by the right-hand rule — exactly the direction produced by the vector (cross) product ω→ × r→. Hence the vec …

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CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/one sentence: Write the relation between angular velocity and time period.$
›Reveal solutionSolution

Angular velocity and time period are related by ω = 2π/T.

For a particle moving in a circle (or any body undergoing uniform rotational motion) with time period T — the time to complete one full revolution — the particle sweeps through an angle of exactly 2π radians (360°) in that time. Since angular velocity is defined as the rate of change of angular displacem …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Dimensional formula of angular velocity is (A) M^0L^0T^-1 (B) ML^-1T^-1 (C) M^0LT^-1 (D) ML^2T^-1$
›Reveal solutionSolution

Angular velocity has dimensional formula M⁰L⁰T⁻¹ because angle itself is dimensionless.

Angular displacement $\theta = \dfrac{\text{arc length}}{\text{radius}} = \dfrac{L}{L}$ , a ratio of two lengths, so $\theta$ is dimensionless: $[\theta] = M^0L^0T^0$ .

…

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CBSE 2025Set sz1 markMCQ
$Q. Linear velocities of all the particles of the rigid body in rotational motion is: (A) One (B) Zero (C) Same (D) Different$
›Reveal solutionSolution

In rotational motion every particle has the same angular velocity, but linear velocity v = omega*r differs because r differs.

When a rigid body rotates about a fixed axis, every particle turns through the same angle in the same time, so all particles have the same angular velocity omega.

The linear (tangential) velocity of a particle is $v = \omega r$ , where r is its perpendicular distance from the axis of rotation. …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The direction of the angular velocity vector is along$
(1) the tangent to the circular path
(2) the inward radius
(3) the outward radius
(4) the axis of rotation
›Reveal solutionSolution

By convention (and the right-hand rule), the angular velocity vector of a rotating body is directed along its axis of rotation, not along the tangent or radius.

For a body rotating about a fixed axis, the angular velocity omega is defined as a vector directed ALONG the axis of rotation (not in the plane of rotation at all). Its sense is given by the right-hand rule: curl the fingers of the right hand in the direction of rotation, and the extended thumb points along omega.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The linear acceleration a of a body rotating about an axis with angular acceleration α, then:$
(a) a = rω
(b) α = ar
(c) a = rα
(d) r = aα
›Reveal solutionSolution

Linear acceleration and angular acceleration are related by $a = r\alpha$ , just as $v=r\omega$ relates linear and angular velocity.

Derivation. For a particle at distance $r$ from the rotation axis, its linear (tangential) velocity is $v = r\omega$ . Differentiating with respect to time (r constant for a rigid body): …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in :$
(a) -z direction
(b) +y direction
(c) -x direction
(d) +z direction
›Reveal solutionSolution

For a particle in uniform circular motion, the angular velocity ω is a vector along the axis of rotation; its direction is fixed by the right-hand rule. Clockwise motion in the xy-plane (viewed from +z) gives ω along -z.

Use the standard right-handed xyz axes: x to the right, y upward, z out of the page toward the viewer. With this convention, a rotation that LOOKS anticlockwise to someone viewing from the +z side corresponds to ω along +z (this is the usual 'anticlockwise = +z' rule you may already know).

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The dimensional formula of angular velocity is —$
(a) [MLT^-1]
(b) [ML^0 T^-1]
(c) [M^0 L^0 T^-1]
(d) [M^0 L^-1 T^-1]
›Reveal solutionSolution

Angular velocity has the dimensional formula [M⁰L⁰T⁻¹].

Angular velocity $\omega = \dfrac{\theta}{t}$ . Angle θ is measured in radians, which is a ratio of two lengths and is therefore dimensionless. …

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CBSE 2021Set ANNUAL1 markMCQ
$Q. The linear velocity v of a body rotating about an axis with angular velocity ω, then :$
(a) v = rα
(b) v = rω
(c) ω = vr
(d) r = vω
›Reveal solutionSolution

Linear speed equals radius times angular speed — this comes directly from the arc-length relation $s = r\theta$ .

For a particle at distance $r$ from the axis of rotation, the arc length swept in angle $\theta$ is $s = r\theta$ . Differentiating with respect to time: …

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CBSE 2021Set TERM11 markMCQ
$Q. A scooter goes round a circular track of radius 10 m with speed of 30 ms^-1. The angular speed of the scooter is:$
(a) 3 rad s^-1
(b) 6 rad s^-1
(c) 300 rad s^-1
(d) None of these
›Reveal solutionSolution

For circular motion, angular speed omega = v/r; with v = 30 m/s and r = 10 m, omega = 3 rad/s.

For a body moving on a circular path of radius r with linear (tangential) speed v, the angular speed omega is defined by:

v = omega x r, so omega = v/r

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in:$
(a) +y direction
(b) +z direction
(c) -z direction
(d) -x direction
›Reveal solutionSolution

Clockwise rotation in the xy-plane corresponds to an angular velocity vector along -z, by the right-hand rule.

For a particle executing uniform circular motion in the xy-plane, the angular velocity ω is a vector directed along the axis of rotation (the z-axis), with its sense given by the right-hand rule: curl the fingers of the right hand in the direction of rotation; the thumb then points along ω.

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. What is the angular displacement made by a particle after 5 s, when it starts from rest with an angular acceleration 0.2 rad s^-2 ?$
(a) 4 rad
(b) 1 rad
(c) 2.5 rad
(d) 5 rad
›Reveal solutionSolution

Starting from rest, the angular displacement after 5 s under angular acceleration 0.2 rad/s^2 is 2.5 rad.

For rotational motion starting from rest (omega0 = 0) with constant angular acceleration alpha, the angular displacement after time t is given by

theta = omega0*t + (1/2)alphat^2

…

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Q.What is the angular displacement made by a particle after 5 s, when it starts from rest with an angular acceleration 0.2 rad s^-2 ?

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