A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that POQ = 0, then the magnitude of the change in velocity is

Key Concepts

1. Uniform circular motion (UCM): Speed $v$ is constant; only direction changes.
2. Velocity as a vector: Even if speed is the same, a new direction means a new velocity vector.
3. Vector subtraction: Change in velocity $\Delta \vec{v} = \vec{v}_Q - \vec{v}_P$.
4. Geometry of equal vectors: Two vectors of equal magnitude making an angle $\theta$ form an isosceles triangle. The side opposite angle $\theta$ gives $|\Delta \vec{v}|$.

Logical Chain

1. At point P, velocity $\vec{v}_P$ is tangential and perpendicular to radius $OP$.
2. At point Q, velocity $\vec{v}_Q$ is tangential and perpendicular to radius $OQ$.
3. Angle between $OP$ and $OQ$ is $\theta$, therefore angle between $\vec{v}_P$ and $\vec{v}_Q$ is also $\theta$.
4. The two vectors $\vec{v}_P$ and $\vec{v}_Q$ have equal length $v$. Place them tail-to-tail; they enclose angle $\theta$.
5. The magnitude of their difference is the closing side of the triangle:

$|\Delta \vec{v}| = 2v\sin\left(\frac{\theta}{2}\right)$

This comes from the Law of Cosines or simple isosceles triangle properties.

Imagine you are riding a bicycle on a round track.
You always ride at the same speed, but when you turn, the direction of your motion changes.
The question asks: “If you ride from one point P to another point Q on this round track and you have turned through an angle $\theta$ at the centre, how much has your velocity (speed + direction) actually changed?”
Because speed stays the same, only the heading changes. We draw two arrows of equal length (speed $v$) pointing tangentially at P and Q; the angle between these arrows is $\theta$. The new arrow that shows ‘how much you had to bend your velocity’ forms the third side of an isosceles triangle. Using simple triangle rules, that “bending” (change in velocity) turns out to be $2v\sin\left(\frac{\theta}{2}\right)$.
So, the sharper you turn (larger $\theta$), the larger the change in velocity.

Given:

• Uniform speed $v$
• Radius of circle $r$
• Central angle swept $\theta$ (in radians)

At point P: velocity vector $\vec{v}_P$ has magnitude $v$.
At point Q: velocity vector $\vec{v}_Q$ has magnitude $v$.
Angle between $\vec{v}_P$ and $\vec{v}_Q$ equals the central angle $\theta$.

Construct an isosceles triangle with sides $\vec{v}_P$ and $\vec{v}_Q$.

Using the Law of Cosines:

$|\Delta \vec{v}|^2 \,=\, v^2 + v^2 - 2v^2\cos\theta \,=\, 2v^2\left(1-\cos\theta\right)$

But $1-\cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$, so

$|\Delta \vec{v}|^2 \,=\, 4v^2\sin^2\left(\frac{\theta}{2}\right)$

Taking square root:

$\boxed{|\Delta \vec{v}| = 2v\sin\left(\dfrac{\theta}{2}\right)}$

Thus, the magnitude of the change in velocity is $2v\sin\left(\dfrac{\theta}{2}\right)$.