Q.A particle rotates in U.C.M. with tangential velocity 'v' along a horizontal circle of diameter 'D'. Total angular displacement of the particle in time 't' is ______. (A) $vt$
(B) $\dfrac{v}{D}t$
(C) $\dfrac{vt}{2D}$
(D) $\dfrac{2vt}{D}$

In rigid-body rotation every particle sweeps the same angle in the same time (same ω), but linear speed v = rω differs with the particle's distance r from the axis.

For a rigid body rotating about a fixed axis, all particles complete the same angular displacement in the same time interval, so every particle has the same angular velocity $\omega$. However, the linear (tangential) speed of a particle at distance $r$ from the …

UCM: constant speed on a circle; velocity direction keeps changing, giving a centripetal acceleration.

Uniform circular motion (UCM) is the motion of a particle along a circular path with constant speed. Although the speed $|v|$ is constant, the direction of velocity changes continuously (it is always tangential to the circle), so the particle has a non-zero acceleration directed towards the centre (centripetal accelera …

In uniform circular motion, speed is constant, so tangential acceleration is zero.

In uniform circular motion (U.C.M.), the speed of the particle stays constant even though its velocity direction keeps changing. Tangential acceleration $a_t = \dfrac{dv}{dt}$ measures the rate of change of speed, which is zero here — only a centripetal (radial) acceleration $a_c = v^2/r$ ex …

Angular displacement $\theta=\omega t$, with $\omega=v/r$ and $r=D/2$.

For a particle in uniform circular motion with tangential (linear) speed $v$ on a circle of diameter $D$ (radius $r=D/2$), the angular velocity is

$\omega=\frac{v}{r}=\frac{v}{D/2}=\frac{2v}{D}$

The total angular displacement in time $t$ is

$\theta=\omega t=\frac{2v}{D}\,t=\frac{2vt}{D}$

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