Angular Kinematics | JEE Notes

All points in a rigid body share the same angular velocity at an instant. They do not share the same linear velocity. A point farther from a fixed axis travels a longer arc in the same time, and therefore moves faster.

Angles, rates and signs

Use radians in mechanics: $θ = s / r$ . A full revolution is $2 π$ radians. The radian is dimensionless, though keeping “rad” visible helps prevent conversion mistakes. Angular displacement can exceed $2 π$ ; it is not restricted to the orientation remaining after complete turns. Define

ω = \frac{d θ}{d t}, α = \frac{d ω}{d t} = \frac{d^{2} θ}{d t^{2}} .

The direction of $α$ describes the change of $ω$ , not necessarily the direction of rotation. A clockwise wheel slowing down has $ω < 0$ and $α > 0$ . It speeds up when $ω α > 0$ , and slows down when $ω α < 0$ . At $ω = 0$ , angular acceleration determines the direction of the impending motion.

Constant acceleration is a condition, not a default

Integrating a constant $α$ once and twice gives

ω = ω_{0} + α t, θ - θ_{0} = ω_{0} t + \frac{12}{α} t^{2}, ω^{2} - ω_{0}^{2} = 2 α (θ - θ_{0}) .

Use these only while $α$ remains constant in the selected signed coordinate. For variable acceleration, retain the derivatives. If $α$ depends on angle, the chain rule gives

α = \frac{d ω}{d θ} \frac{d θ}{d t} = ω \frac{d ω}{d θ}, \frac{12}{(} ω^{2} - ω_{0}^{2}) = \int_{θ_{0}}^{θ} α (ϑ) d ϑ .

If $α$ depends on time, integrate $ω = ω_{0} + \int α (t) d t$ , then integrate $ω$ for angle. If it depends on $ω$ , separate $d ω / α (ω) = d t$ . Choose the representation matching the given dependence; memorising additional special-case formulas is unnecessary.

Graph	Slope	Signed area	Frequent misreading
$θ$ against $t$	$ω$	No standard basic kinematic meaning	A negative slope means clockwise motion, not necessarily slowing
$ω$ against $t$	$α$	Angular displacement	Total angle travelled uses $\int \| ω \| d t$
$α$ against $t$	Rate of change of angular acceleration	Change of angular velocity	Zero area means equal final and initial $ω$ , not no motion
$ω^{2}$ against $θ$	$2 α$	No standard basic kinematic meaning	Squaring hides the direction of rotation

Watch the animation: Rolling is slide plus spin

Worked example 03 · Acceleration specified by angle

Problem. A rotor starts with $ω_{0} = 2 r a d s^{- 1}$ at $θ = 0$ . While it turns positively, $α = k θ$ , where $k = 3 s^{- 2}$ . Find its angular speed after an angular displacement of $2$ radians.

Step 1: choose angle as the variable. $ω d ω = k θ d θ$ . Direct use of $ω = ω_{0} + α t$ would incorrectly treat a changing acceleration as constant.

Step 2: integrate with limits.

\int_{ω_{0}}^{ω} ω^{'} d ω^{'} = \int_{0}^{θ} k ϑ d ϑ \Rightarrow ω^{2} = ω_{0}^{2} + k θ^{2} .

Step 3: select the physical branch. Rotation starts positive and $α \geq 0$ on the stated path, so $ω = \sqrt{4 + 3 (2)^{2}} = 4 r a d s^{- 1}$ .

Checks. Since radians are dimensionless, $k θ^{2}$ has units $s^{- 2}$ , matching $ω^{2}$ . As $k \to 0$ , the initial speed is retained. The answer is a speed; a squared equation alone cannot determine a sign in a reversal problem.

Velocity and acceleration of a point on the body

Let $C$ and $P$ be material points fixed in the same rigid body, with $𝐫 = 𝐫_{P} - 𝐫_{C}$ . The reference point $C$ may be the centre of mass, but need not be. Rigidity means $𝐫$ changes direction while retaining its length. Its inertial derivative is $d 𝐫 / d t = 𝛚 \times 𝐫$ . Therefore,

𝐯_{P} = 𝐯_{C} + 𝛚 \times 𝐫,

𝐚_{P} = 𝐚_{C} + 𝛂 \times 𝐫 + 𝛚 \times (𝛚 \times 𝐫) .

For planar motion, the last term is $- ω^{2} 𝐫$ , directed toward the reference point relative to which the rotation is described. Relative tangential acceleration has magnitude $| α | r$ , and relative normal acceleration has magnitude $ω^{2} r$ . These are perpendicular, so for a fixed reference point on the axis, $a = \sqrt{(α r)^{2} + (ω^{2} r)^{2}}$ . If the reference point accelerates, include $𝐚_{C}$ vectorially before calculating the magnitude.

Uniform angular speed does not mean zero acceleration: a point moving in a circle constantly changes velocity direction. Conversely, at the instant $ω = 0$ , the centripetal term vanishes while the tangential term can remain nonzero. These formulas apply to points attached to the body; a bead sliding along a rotating rod requires additional relative-motion terms.

Instantaneous centre: a geometric shortcut with a strict boundary

When $ω \neq 0$ , planar rigid motion has an instantaneous point I with zero velocity. Then $𝐯_{P} = 𝛚 \times 𝐫_{P / I}$ , so each point’s velocity is perpendicular to the line joining it to I. If two point-velocity directions are known, draw perpendiculars to those directions through the points; their intersection locates I. Speed ratios become distance ratios, $v_{A} / v_{B} = I A / I B$ . This can solve a rod sliding between a wall and floor without differentiating a trigonometric constraint.

Watch the animation: Sliding ladder and the instantaneous centre

The shortcut describes velocities at one instant. I may lie outside the body and may move to a different location in the next instant. Do not automatically set its acceleration to zero or treat it as a permanent hinge in a torque equation. For pure translation, $ω = 0$ , there is no finite instantaneous centre; equal velocities are better handled directly.

Low yield for JEE: belts and gears

Worked example 04 · An open belt followed by a gear pair

Problem. An open, inextensible belt without slip connects pulleys A and B of radii $0.10$ m and $0.25$ m. A rotates counterclockwise at $15 r a d s^{- 1}$ , with angular acceleration $4 r a d s^{- 2}$ . A gear C of pitch radius $0.08$ m is rigidly attached to B and meshes externally with a gear D of pitch radius $0.20$ m. Find $ω_{D}$ and $α_{D}$ .

Step 1: apply belt compatibility. An open belt makes both pulleys rotate in the same sense. Matching tangential speeds gives $r_{A} ω_{A} = r_{B} ω_{B}$ , hence $ω_{B} = 6 r a d s^{- 1}$ . Differentiating gives $r_{A} α_{A} = r_{B} α_{B}$ , so $α_{B} = 1.6 r a d s^{- 2}$ .

Step 2: use the common axle. $ω_{C} = ω_{B}$ and $α_{C} = α_{B}$ . Equal angular velocities do not require equal radii.

Step 3: apply external gear compatibility. The gears turn oppositely: $r_{C} ω_{C} + r_{D} ω_{D} = 0$ . Thus $ω_{D} = - 2.4 r a d s^{- 1}$ and $α_{D} = - 0.64 r a d s^{- 2}$ .

Checks. Radius ratios are dimensionless, so units are preserved. Larger driven radii reduce angular speed. A crossed belt would reverse the A-to-B sign. Match tangential accelerations through these constraints; do not equate complete acceleration vectors at different pulley rims.

Worked example 05 · An instantaneously stationary point still accelerates

Problem. At one instant, a wheel of radius $R$ rolls right without slipping on a stationary horizontal surface. Its centre has $𝐯_{C} = v \hat{𝐱}$ , $𝐚_{C} = a \hat{𝐱}$ , and the rolling constraints give $𝛚 = - (v / R) \hat{𝐳}$ , $𝛂 = - (a / R) \hat{𝐳}$ . Find the velocity and acceleration of the material point P currently touching the ground.

Step 1: set the position vector. $𝐫_{P / C} = - R \hat{𝐲}$ .

Step 2: calculate velocity. $𝛚 \times 𝐫_{P / C} = - v \hat{𝐱}$ , so $𝐯_{P} = 𝟎$ .

Step 3: calculate acceleration. The tangential term is $- a \hat{𝐱}$ , cancelling $𝐚_{C}$ . The centripetal term is $+ ω^{2} R \hat{𝐲}$ . Hence

𝐚_{P} = \frac{v^{2}}{R} \hat{𝐲} .

Checks. $v^{2} / R$ has acceleration units and vanishes when $v = 0$ . The point is stationary only at that instant; immediately afterward it rises away from the ground. The contact location continuously changes from one material point to another. An instantaneous centre of zero velocity is not automatically a fixed pivot with zero acceleration.

Checkpoint · A wheel has

ω = - 6 r a d s^{- 1}

and

α = + 2 r a d s^{- 2}

, constant. What happens over four seconds?

It slows clockwise, stops after $3$ s, then accelerates counterclockwise. Net displacement is $- 6 (4) + \frac{12}{(} 2) (4)^{2} = - 8$ rad. Angle travelled is $9 + 1 = 10$ rad: split the motion at the reversal. Final angular velocity is $+ 2 r a d s^{- 1}$ .

Two quick results JEE reuses

Direction of net acceleration. From rest with constant $α$ , a rim point has $a_{t} = r α$ and $a_{r} = r ω^{2} = 2 r α θ$ . The net acceleration is at $45^{\circ}$ to the velocity when $a_{r} = a_{t}$ , so at $θ = \frac{12}{}$ rad, whatever the radius.

Sliding ladder. Draw a normal to each end's velocity. They meet at the instantaneous centre $I$ , which is the corner of the rectangle formed with the wall and floor. Then $ω = v_{t o p} / x = v_{f o o t} / y$ , and $v_{f o o t} / v_{t o p} = y / x$ . The midpoint stays at $L / 2$ from the corner, so it moves on a circle.

Angular kinematics and the motion of every point

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