Circular Motion and Centripetal Force

Why circular motion needs a force

Velocity is a vector. A body moving in a circle keeps constant speed only if its direction keeps changing — and direction change is a change of velocity, which is acceleration, which by Newton's second law requires a net force. A body in circular motion is never in equilibrium, even when the speedometer reads steady.

For a body on a circle of radius \(R\) at speed \(v\), the acceleration points from the body toward the centre, with magnitude

\[ a_c = \frac{v^2}{R} = \omega^2 R. \]

This is the centripetal acceleration — a kinematic result derived in the Chapter 3 sibling on uniform circular motion. Our job here is the dynamics: identifying which real force produces that acceleration.

Centripetal force is a role, not a new force

NCERT states the dynamics directly:

"\(F_c = m v^2/R\), directed toward the centre, is called the centripetal force."

The key word is "called". Centripetal force is not an extra entry on your force inventory. It is the name we give to the net inward force in circular motion, once Newton's second law is applied along the radial axis. The actual provider is always one of the standard forces you already know — tension, gravity, normal contact, friction, electrostatic attraction. Identifying the provider in each problem is what NEET is really testing.

Bridge from the kinematics article

Taking \(a_c = v^2/R\) as a black box, Newton's second law gives the inward equation

\[ \sum F_{\text{inward}} = m\,a_c = \frac{m v^2}{R} = m\omega^2 R. \]

The other equation you almost always need is the perpendicular balance — usually vertical balance, because most NEET setups have gravity downward and a normal or vertical tension component balancing it. Two equations, two unknowns, single-unknown chain.

The provider inventory NEET tests

Across NCERT examples and NEET PYQs, the centripetal force is supplied by one of five providers. Recognising the provider is half the marks.

Every problem on circular dynamics reduces to picking the right row and pairing it with the perpendicular balance equation.

Car on a level road — full FBD

A car of mass \(m\) drives at speed \(v\) around a horizontal circular bend of radius \(R\). The road is flat — no banking. Three real forces act on the car:

1. Weight \(mg\) downward.
2. Normal force \(N\) upward.
3. Friction \(f\) horizontal, toward the centre of the turn.

Vertical balance: \(N = mg\). Horizontal balance is the centripetal equation — friction is the only horizontal force, so

\[ f = \frac{m v^2}{R}. \]

Friction has a ceiling: \(f \le \mu_s N = \mu_s mg\). Therefore

\[ \frac{m v^2}{R} \le \mu_s\, m g \;\Longrightarrow\; v^2 \le \mu_s R g. \]

The maximum safe cornering speed on a level road is

\[ \boxed{\,v_{\max} = \sqrt{\mu_s R g}\,.} \]

The mass has cancelled. A loaded truck and an empty hatchback share the same level-road speed limit on the same curve, provided \(\mu_s\) and \(R\) are the same.

For larger speeds, engineers bank the road — tilt it inward so that the normal force gains a horizontal component and shares the centripetal load with friction. The full \(v_{\max} = \sqrt{R g\,(\mu_s + \tan\theta)/(1 - \mu_s \tan\theta)}\) derivation is in the next sibling article on banking of roads.

Conical pendulum derivation

A bob of mass \(m\) hangs from a light string of length \(L\). It is set into uniform circular motion in a horizontal plane so the string traces a cone of half-angle \(\theta\) with the vertical. The radius of the horizontal circle is \(r = L \sin\theta\).

Only two real forces act: gravity \(mg\) downward and tension \(T\) along the string toward the pivot. Decompose tension. Vertical component balances gravity:

\[ T \cos\theta = m g. \]

Horizontal component supplies centripetal force:

\[ T \sin\theta = \frac{m v^2}{r}. \]

Divide:

\[ \tan\theta = \frac{v^2}{r g}. \]

The speed at a given cone angle is \(v = \sqrt{r g \tan\theta}\), and with \(r = L\sin\theta\) the period is

\[ T_{\text{period}} = 2\pi\sqrt{\frac{L \cos\theta}{g}}. \]

Vertical circle: top and bottom

Now swing the stone in a vertical plane. Gravity sometimes helps tension provide centripetal force and sometimes opposes it; speed varies around the loop.

At the top of the loop

At the top, both \(T_{\text{top}}\) and weight \(mg\) point downward — straight toward the centre. The radial equation is

\[ T_{\text{top}} + m g = \frac{m v_{\text{top}}^2}{R}, \qquad T_{\text{top}} = \frac{m v_{\text{top}}^2}{R} - m g. \]

A string can only pull, so \(T_{\text{top}} \ge 0\). The borderline \(T_{\text{top}} = 0\) gives

\[ \boxed{\,v_{\text{top, min}} = \sqrt{g R}\,.} \]

At this speed gravity alone supplies the centripetal force and the string just goes slack at the top.

At the bottom of the loop

At the bottom, \(T_{\text{bot}}\) points upward (toward centre) and \(mg\) points downward (away from centre):

\[ T_{\text{bot}} - m g = \frac{m v_{\text{bot}}^2}{R}. \]

Linking top and bottom by energy conservation

Tension is perpendicular to velocity and does no work; only gravity does work. Between bottom and top the stone rises by \(2R\), so

\[ v_{\text{bot}}^2 = v_{\text{top}}^2 + 4 g R. \]

Substituting \(v_{\text{top}}^2 = g R\):

\[ v_{\text{bot, min}}^2 = 5 g R \;\Longrightarrow\; \boxed{\,v_{\text{bot, min}} = \sqrt{5 g R}\,.} \]

Roller-coaster loops and the "rotor" ride satisfy the same condition, with the seat or track replacing the string.

Worked examples