Q.For a point on a rotating rigid body, the graph of its angular position θ (plotted on the vertical axis) against time t (plotted on the horizontal axis) is a straight line of constant positive slope: θ increases uniformly with t, passing steadily through successive instants t1<t2<t3. Is the body rotating clockwise or anti-clockwise? Give the reason.
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 = 2π 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 θ(t) — the angle it has turned through from some reference line. Then:
ω=dtdθ
where ω (Greek letter omega) is the instantaneous angular velocity. For uniform rotation (constant speed), this simplifies to:
ω=ΔtΔθ
Units: radians per second (rad/s). In practice, you'll also see revolutions per minute (rpm) — 1 rpm = 602π 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 ω.
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=ωr
This is why the outer edge of a merry-go-round moves faster than a point near the centre — same ω, different r.
Watch out
This formula v=ωr only works when v is the tangential speed (perpendicular to the radius). It does NOT apply to radial motion (straight in or out).
The straight θ-t line has a constant positive slope, so the angular velocity ω=dθ/dt is positive and constant. Increasing angular position corresponds, by the usual sign convention, to anti-clockwise rotation. …
A straight-line θ versus t graph means θ changes at a constant rate, so the angular velocity ω=dθ/dt is constant. Its slope here is positive, so θ is increasing. By the standard convention that increasing angular position is measured anti-clockwise, the body turns anti-clockwise.
Concept
Angular velocity is the slope of the angular-position–time graph: ω=dtdθ.
Reasoning
The graph is a straight line, so dtdθ is constant — the rotation is uniform.
The slope is positive, so ω>0; the angular position θ keeps increasing with time. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2026Set 2026-M1 markMCQ
Q.A particle starts rotating from rest. The instantaneous angular displacement is θ=3t3−t2, where θ is in radian and t in s; The angular velocity at t=1s is
(A) 9rads−1
(B) 7rads−1
(C) 3rads−1
(D) 1rads−1
›Reveal solutionSolution
The angular velocity is the first derivative of angular displacement with respect to time. Differentiating θ=3t3−t2 gives ω=9t2−2t, and at t=1 s this equals 7 rad/s. The correct option is (B).
Concept & Intuition
Angular velocity ω tells us how fast the angle θ is changing at a given instant. When you have an equation for θ(t), the instantaneous rate of change is found by differentiating with respect to time — just like how velocity in a straight line is the derivative of position. Here, the particle starts from rest, but that only tells us ω(0)=0; we need ω(1).
Step-by-step solution
Recall the definition
Instantaneous angular velocity is the derivative of angular displacement:
ω=dtdθ
Differentiate the given function
Given θ=3t3−t2, differentiate term by term:
Q.A particle is in uniform circular motion. The equation of its trajectory is given by (x−2)2+y2=25, where x and y are in meter. The speed of the particle is 2ms−1, when the particle attains the lowest ‘y’ co-ordinate, the acceleration of the particle is (in ms−2)
(A) 0.4j^
(B) 0.8i^
(C) 0.8j^
(D) 0.4i^
›Reveal solutionSolution
Read the centre and radius off the circle's equation, locate the lowest point, then apply a=rv2n^ with n^ pointing from the particle to the centre.
Step 1 — Identify the circle.
Compare
(x−2)2+y2=25
with the standard form (x−h)2+(y−k)2=R2:
Centre C=(2,0),R=25=5m
Step 2 — Locate the point of interest.
On this circle y ranges over [−5,+5]. The lowest y-coordinate is y=−5, attained where (x−2)2=0, i.e. at
P=(2,−5)
This is the bottom of the circle, directly below the centre.
Step 3 — Acceleration in uniform circular motion.
"Uniform" means the speed is constant, so the tangential acceleration is zero:
at=dtdv=0
The entire acceleration is therefore centripetal — it changes only the direction of v, never its magnitude. Its magnitude is
ac=Rv2
and it always points from the particle towards the centre.