Animated Solution for Physics - Kinematics: A particle is moving along a circular path with a constant speed of $10{\text{ ms}}^{-1}$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{\circ }$ around the centre of the circle?

When a particle moves in a circle at a constant speed, it is tempting to think that its velocity isn't changing. However, velocity is a vector quantity, meaning it possesses both magnitude (speed) and direction. Even if the speedometer reads a steady $10{\text{ ms}}^{-1}$, the steering wheel is constantly turning. Because the direction of motion is continuously altering, the velocity is continuously changing.

In this problem, we are asked to find the magnitude of this change, denoted as $|\Delta \vec{v}|$, after the particle has swept through an angle of $60^{\circ }$.

Let's visualize the setup. The particle starts at point $A$ with an initial velocity ${\vec{v}}_1$. After traveling along the arc, it reaches point $B$ with a final velocity ${\vec{v}}_2$. Both vectors are tangent to the circular path at their respective points.

If we take these two velocity vectors and place them tail-to-tail, what is the angle between them? A beautiful property of circular motion is that the velocity vector is always perpendicular to the position vector (the radius). Therefore, if the radius rotates by $60^{\circ }$, the perpendicular velocity vector must also rotate by exactly $60^{\circ }$. The angle between ${\vec{v}}_1$ and ${\vec{v}}_2$ is exactly $60^{\circ }$.

We need to find $\Delta \vec{v}={\vec{v}}_2-{\vec{v}}_1$. Geometrically, if we draw ${\vec{v}}_1$ and ${\vec{v}}_2$ tail-to-tail, the vector $\Delta \vec{v}$ is the line connecting the head of ${\vec{v}}_1$ to the head of ${\vec{v}}_2$.

Let's analyze the triangle formed by these three vectors. We know two things: 1. The magnitudes of the initial and final velocities are equal: $|{\vec{v}}_1|=|{\vec{v}}_2|=10{\text{ ms}}^{-1}$. This makes it an isosceles triangle. 2. The angle between these two equal sides is $60^{\circ }$.

An isosceles triangle with an included angle of $60^{\circ }$ forces the other two angles to also be $60^{\circ }$ (since $(180^{\circ }-60^{\circ })/2=60^{\circ }$). This means our vector triangle is actually an equilateral triangle!

Because all sides of an equilateral triangle are equal, the magnitude of the change in velocity must equal the initial speed:

While the equilateral triangle trick is elegant for $60^{\circ }$, what if the angle was different? We can use the law of cosines for vector subtraction:

Substituting $|{\vec{v}}_1|=|{\vec{v}}_2|=v$:

Using the half-angle trigonometric identity $1-\cos \theta =2{\sin }^2(\frac{\theta }{2})$:

If you plug $\theta =60^{\circ }$ and $v=10$ into this master formula, you will arrive at the exact same result: $2(10)\sin (30^{\circ })=20(0.5)=10{\text{ ms}}^{-1}$.