Angular Displacement Formula

Angular displacement is the angle measured in radians and is defined as the shortest angle between the initial and the final points for a given object undergoing circular motion about a fixed point. Angular Displacement is a vector quantity. It has both direction and magnitude. It is represented by a circular arrow pointing from the initial point to the final point, which is either clockwise or anti-clockwise in direction.

Formula of Angular Displacement

Angular displacement of a point can be given by using the following formula,

\(\begin{array}{l}Angular  displacement = \theta _{f}- \theta _{i}\end{array} \)


\(\begin{array}{l}\theta = s/r\end{array} \)

Here, θ is the angular displacement of the object through which the movement has occurred, s is the distance covered by the object on the circular path and r is the radius of curvature of the given path.  

When the acceleration of the object (α), the initial angular velocity (ω) and the time (t) at which the displacement is to be calculated is known, we can use the following formula.

\(\begin{array}{l}\theta = wt + 1/2 \alpha t^{2}\end{array} \)

Derivation of Angular Displacement Formula

Let us consider an object ‘A’ undergoing linear motion with initial velocity ‘u’ and acceleration ‘a’. Let us say, after time t, the final velocity of the object is ‘v’ and the total displacement of the object is ‘s’.

We know that acceleration is defined as the rate of change of velocity. Therefore,

\(\begin{array}{l}a = \frac{\mathrm{d} v}{\mathrm{d} t}\end{array} \)

\(\begin{array}{l}dv = a dt\end{array} \)

Integrating on both sides,

\(\begin{array}{l}\int_{u}^{v} dv = a \int_{0}^{t} dt\end{array} \)

\(\begin{array}{l}v – u = at\end{array} \)


\(\begin{array}{l}a = \frac{\mathrm{d} v}{\mathrm{d} t}\end{array} \)

\(\begin{array}{l}a = \frac{\mathrm{d} v}{\mathrm{d} x} *\frac{\mathrm{d} x}{\mathrm{d} t}\end{array} \)

since  v=dx/dt, we can write,

\(\begin{array}{l}a = v \frac{\mathrm{d} v}{\mathrm{d} x}\end{array} \)

v dv=a dx

Upon integrating both the sides,we get

\(\begin{array}{l}\int_{u}^{v} v dv = a \int_{0}^{s} dx\end{array} \)

\(\begin{array}{l}v^{2} – u^{2} =2as\end{array} \)

Substituting the value of u from the equation 1 into the second equation, we get,

\(\begin{array}{l}v^{2} -(v- at)^{2} = 2as\end{array} \)

\(\begin{array}{l}2vat – a^{2}t^{2}= 2as\end{array} \)

Dividing both the sides of the equation by 2a, we get,

\(\begin{array}{l}s = vt – \frac{1}{2} at^{2}\end{array} \)

Upon substituting the value of v instead of u we get,

\(\begin{array}{l}s = ut + \frac{1}{2}at^{2}\end{array} \)


Problem 1:

1) Neena goes around a circular track that has a diameter of 7 m. If she runs around the entire track for a distance of 50 m, what is her angular displacement?


According to question, Neena’s linear displacement, s = 50 m.

Also, the diameter of the curved path, d = 7 m

As we know that, d = 2r, so r =7/2= 3.5 m

And according to the formula for angular displacement,

\(\begin{array}{l}\theta = \frac{s}{r}\end{array} \)

θ = 50m /3.5 m

θ = 14.28 radians

Problem 2

2) Rohit bought a pizza of a radius of 0.5 m. A fly lands on the pizza and walks around the edge for a distance of 80 cm. Calculate the angular displacement of the fly?


According to the question, the distance traveled by the fly on the pizza is s = 80 cm = 0.08 m.

The radius of the pizza is given to be, r = 0.5 m.

Using the formula for angular displacement,

\(\begin{array}{l}\theta = \frac{s}{r}\end{array} \)

θ = 0.08m/0.5 m

θ = 0.16 radians


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