Circular Motion - Quick Reference

Angular Measurements and Displacement

Circular motion is described by angular position and displacement. Angular displacement Δθ is the angle swept by the radius vector and is measured in radians. Arc length s and radius r are related by $s = r\,\Delta\theta$. Angles measured clockwise (CW) are negative and those measured counterclockwise (CCW) are positive. A full revolution corresponds to $2\pi\text{ rad} = 360^\circ$. The angular position is denoted by Θ. The direction of the angular displacement vector is given by the right‑hand rule: curl the fingers in the direction of rotation and the thumb points along the angular displacement vector.

Uniform Circular Motion: Velocity, Acceleration and Force

In Uniform Circular Motion (UCM) the speed is constant but the velocity vector changes direction. The velocity is tangent to the circle and has magnitude $v$. The centripetal (radial) acceleration is $a<em>c = \dfrac{v^2}{r} = \omega^2 r$, directed toward the center. By Newton’s second law, the required centripetal force is $F</em>c = m a_c = m \dfrac{v^2}{r}$, acting along the radius toward the center.

Velocity and Acceleration in UCM

The tangential component of acceleration relates to angular acceleration by $a<em>t = r\,\alpha$. For UCM, the angular acceleration is zero ((\alpha = 0)), so the acceleration is purely radial: $a = a</em>r = \dfrac{v^2}{r} = \omega^2 r$. The instantaneous velocity is magnitude $v = r\,\omega$ and is tangent to the path.

Right-Hand Rule and Direction of Vectors

The direction of the angular displacement (and angular velocity) is given by the right-hand rule: curling the fingers in the direction of rotation, the thumb points along the angular velocity vector.

Centripetal Force: Physical Meaning

Centripetal force is a real force that acts along the radius toward the center, enabling circular motion. In UCM, the required force is provided by a real interaction (grip in a string, tension, gravity components, friction, normal force, etc.). The radial (centripetal) acceleration magnitude is $a_c = \dfrac{v^2}{r}$.

Banked Roads and Friction

A car moving along a curved road in circular motion can rely on the horizontal component of the contact forces to provide the centripetal force. For a frictionless banked curve, the relation is $\tan\theta = \dfrac{v^2}{g\,r}$, so the maximum safe speed is $v^2 = g\,r\,\tan\theta$.

When friction is present, the maximum safe speed becomes
$v<em>{\max}^2 = \dfrac{r g (\tan\theta + \mu</em>s)}{1 - \mu<em>s \tan\theta}$ where (\mus) is the coefficient of static friction. If friction is insufficient, skidding occurs. The normal reaction (N) and frictional force (f) can be resolved into vertical and horizontal components to balance weight and provide the centripetal force.

Key implications:

• Banking helps provide CPF and can reduce tyre wear; mass cancels from the banking angle (for the basic no‑friction banking case).
• Friction can supplement or replace banking to provide CPF; the frictional limit imposes a speed limit on the curve.

Conical Pendulum: Time Period

A conical pendulum consists of a bob of mass (m) on a string of length (\ell), the bob describing a horizontal circle of radius (r) while the string makes a cone angle (\theta) with the vertical. Forces: weight (mg) downward, tension (T) along the string. Resolve tensions:

• Vertical balance: $T\cos\theta = mg$
• Horizontal (radial) component provides centripetal force: $T\sin\theta = \dfrac{m v^2}{r}$
  From these, $\tan\theta = \dfrac{v^2}{g\,r}$ and hence $v^2 = g\,r\,\tan\theta$.
  With (r = \ell\sin\theta) and (h = \ell\cos\theta) (vertical height), we have (\tan\theta = \dfrac{r}{h}) and thus
  $v^2 = g\,r\tan\theta = g\,\dfrac{r^2}{h}$.
  The period is
  $T_{p} = \dfrac{2\pi r}{v} = \dfrac{2\pi r}{\sqrt{g r^2/h}} = 2\pi\sqrt{\dfrac{h}{g}}$
  So the period depends on height (h) and gravity, but not on mass or string length.

Vertical Circular Motion Under Gravity

Consider a mass (m) on a string of radius (r) performing vertical circular motion. At the top (point A) with speed (v1): $T</em>1 + mg = \dfrac{m v<em>1^2}{r}$ If the top velocity is the minimum needed to keep the string taut, then (T1 = 0) and
$v<em>1^2 = gr$ At the bottom (point B) with speed (v2):
$T<em>2 - mg = \dfrac{m v</em>2^2}{r}$
Using energy conservation between top and bottom, the change in potential energy is (2 r\,m g), giving
$\tfrac{1}{2} m v<em>2^2 = \tfrac{1}{2} m v</em>1^2 + 2 m g r \quad\Rightarrow\quad v<em>2^2 = v</em>1^2 + 4 g r$
Combining with (v1^2 = gr) yields $v</em>2^2 = 5 g r\,,$
so the speed is largest at the bottom. The energy (kinetic plus potential) is constant along the path:
$\tfrac{1}{2} m v^2 + m g y = \text{constant}.$
The so‑called critical or minimum top velocity to maintain a taut string is (v_1 = \sqrt{gr}).

Quick Reference: Key Equations

• Arc length and angle: $s = r\Delta\theta,\quad 2\pi\text{ rad} = 360^\circ$
• Uniform circular motion: $v = r\omega,\quad a<em>c = \dfrac{v^2}{r} = \omega^2 r,\quad F</em>c = m\dfrac{v^2}{r}$
• Tangential vs radial acceleration: $a<em>t = r\alpha,\quad a = \sqrt{a</em>t^2 + a<em>r^2},\quad a</em>r = \dfrac{v^2}{r},\ a_t = r\alpha$
• Banked road (no friction): $\tan\theta = \dfrac{v^2}{g r},\quad v^2 = g r \tan\theta$
• Banked road with friction: $v<em>{\max}^2 = \dfrac{r g (\tan\theta + \mu</em>s)}{1 - \mu_s \tan\theta}$
• Conical pendulum: $T\cos\theta = mg,\quad T\sin\theta = \dfrac{m v^2}{r},\quad v^2 = g r \tan\theta$
• Period of conical pendulum: $T_p = \dfrac{2\pi r}{v} = 2\pi\sqrt{\dfrac{h}{g}}\quad(\text{with } r/h = \tan\theta)$
• Vertical circular motion: top, minimum taut velocity $v<em>1^2 = gr,\quad v</em>2^2 = v_1^2 + 4gr = 5gr$
• Energy conservation: $\tfrac{1}{2} m v^2 + m g y = \text{constant}$