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.
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🔒 Start your 14-day free trial to unlock the full solution →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).
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:
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.
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).
A Concrete Example …
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. …
Concept: Angular Velocity Is the Slope of the θ–t Graph
ω=dtdθ
Step 1: Read the shape of the graph
The graph is a straight line, so dθ/dt is constant — the body rotates uniformly (constant ω).
Step 2: Read the sign of the slope
The slope is positive, so ω>0: the angular position θ keeps increasing as t increases through t1<t2<t3.
Step 3: Apply the standard sign convention …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.Ratio of angular velocity of hour hand of a watch and the angular velocity of rotation of earth is (A) 1 : 1 (B) 2 : 1 (C) 4 : 1 (D) 1 : 2
›Reveal solutionSolution
Angular velocity is inversely proportional to time period; comparing the 12-hour period of the hour hand to Earth's 24-hour rotation period gives a ratio of (B) 2 : 1.
Concept and Intuition
Angular velocity is defined as ω=2π/T where T is the time period for one complete revolution. A shorter period means a larger (faster) angular velocity. The hour hand of a 12-hour analogue watch sweeps through 360∘ once every 12 hours, while the Earth completes one full rotation about its own axis once every 24 hours (taking the everyday approximation of a 24-hour day, as is standard for this kind of ratio problem).
Step-by-Step Solution
- Time period of hour hand: Thour=12 hours.
- Time period of Earth's rotation about its axis: Tearth=24 hours.
- Angular velocities: ωhour=122π, ωearth=242π. …
- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.The angular velocity of a circular disc rotating with uniform angular acceleration increases from 20π rad s−1 to 50π rad s−1 in a time of 10 seconds. The number of rotations made by the circular disc during this period is (A) 125 (B) 150 (C) 175 (D) 200
›Reveal solutionSolution
Using uniform angular acceleration kinematics, the disc sweeps 350π radians in 10 s, i.e. 175 complete rotations.
Concept and Intuition
For uniformly accelerated rotational motion, angular kinematics mirror linear kinematics: θ=ω1t+21αt2 (or equivalently, since acceleration is uniform, θ=(2ω1+ω2)t, using the average angular velocity). Once the total angle turned is known, dividing by 2π gives the number of complete revolutions.
Step-by-Step Solution
- Find angular acceleration: α=tω2−ω1=1050π−20π=1030π=3π rads−2.
- Total angle: θ=ω1t+21αt2=(20π)(10)+21(3π)(10)2=200π+150π=350π rad. …
- AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQQ.A particle revolving in a circular path travels the first half of the circumference in 4 s and the next half in 2 s. What is its average angular velocity? (A) 4π/9 rad/s (B) π/6 rad/s (C) 2π/3 rad/s (D) π/3 rad/s
›Reveal solutionSolution
Average angular velocity only needs the total angle and total time, not the two different half-times separately. Answer: π/3 rad/s.
Concept and Intuition
Average angular velocity is defined as total angular displacement divided by total time elapsed — it does not require knowing the instantaneous angular velocity at every moment, so the fact that the two halves took different times doesn't need any weighted-average trick.
Step-by-Step Solution
- The particle completes one full circle, sweeping through 2π radians total.
- Total time taken =4s+2s=6s.
- ωavg=ΔtΔϕ=62π=3π rad/s. …
- AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQQ.An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 seconds. The linear speed of the insect is ______ (A) 4.3 cm.s−1 (B) 5.3 cm.s−1 (C) 6.3 cm.s−1 (D) 7.3 cm.s−1
›Reveal solutionSolution
This tests converting revolutions-per-time into linear speed using the circular path's circumference.
Concept and Intuition
For an object moving steadily around a circular path, the total distance travelled equals the number of complete revolutions multiplied by the circumference of the circle (2πr). Dividing this distance by the total time gives the (constant) linear/tangential speed along the groove.
Step-by-Step Solution
- Radius r=12 cm, so circumference =2πr=2×3.1416×12=75.40 cm.
- Number of revolutions n=7 in time t=100 s.
- Total distance travelled =n×2πr=7×75.40=527.79 cm. …
- AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQQ.The angular speed of a fly-wheel making 180 rpm is ________ (A) (4π) rad.s−1 (B) (6π) rad.s−1 (C) (2π) rad.s−1 (D) (π54) rad.s−1
›Reveal solutionSolution
Converting rpm to rad/s using ω=2πf, with f in revolutions per second, gives 6π rad/s.
Concept and Intuition
Angular speed ω (in rad/s) relates to rotational frequency f (in revolutions per second) by ω=2πf, since one full revolution corresponds to an angle of 2π radians. Revolutions per minute (rpm) must first be converted to revolutions per second by dividing by 60.
Step-by-Step Solution
- Convert rpm to rps: f=60180 rpm=3 rev/s.
- Apply ω=2πf=2π(3)=6π rad.s−1. …
- AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQQ.If ω=3i^−5j^+2k^ and r=5i^−6j^+6k^, then find linear velocity ________ (A) −18i^−8j^+7k^ (B) −18i^−13j^+2k^ (C) 4i^−13j^+6k^ (D) 6i^−2j^+8k^
›Reveal solutionSolution
This tests the vector relation v=ω×r for a rotating rigid body; the determinant cross-product evaluates to −18i^−8j^+7k^.
Concept and Intuition
For a particle at position vector r in a body rotating with angular velocity ω, the linear (tangential) velocity of that particle is given by the vector cross product v=ω×r. This captures both the magnitude (ωrsinθ) and the direction (perpendicular to both ω and r, following the right-hand rule) of the particle's velocity.
Step-by-Step Solution
- Write ω=3i^−5j^+2k^ and r=5i^−6j^+6k^.
- Set up the determinant: v=i^35j^−5−6k^26.
- i^ component: (−5)(6)−(2)(−6)=−30+12=−18. …
- AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQQ.A particle moves in a circle of radius 5 m with a linear velocity of 25 m.s−1. Its angular velocity is ________ (A) 5 rad.s−1 (B) 0.2 rad.s−1 (C) 10 rad.s−1 (D) 0.1 rad.s−1
›Reveal solutionSolution
A direct application of the linear-angular velocity relation v=ωr gives ω=5 rad.s−1.
Concept and Intuition
For uniform circular motion, the linear (tangential) speed v of a particle relates to its angular speed ω and the radius r of its circular path through v=ωr. This comes from the arc-length relation s=rθ, differentiated with respect to time.
Step-by-Step Solution
- Given: radius r=5 m, linear velocity v=25 m.s−1. …
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