MHT-CET Physics · Rotational Dynamics

Dynamics of Circular Motion — Banking, Conical Pendulum and the Vertical Circle

Something must supply the centripetal force mv²/r — a string, a spring, friction, the normal reaction of a banked road or a funnel — and resolving that force into vertical and horizontal parts answers every question here.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

17 PYQs, three HARD. Three shapes: what the centripetal force is and how it scales, a force at an angle (banked road, conical pendulum, a bob hanging in a turning car), and the vertical circle, where speed and tension change from top to bottom.

Concept 1 of 3: The Centripetal Force and What Supplies It

Centripetal force is not a new force; it is the job some real force is doing — the tension in a string, a spring's pull, the weight of a hanging mass fed through a hole. Write that force equal to mv²/r and solve.

Definition

$F = \dfrac{mv^2}{r} = m\omega^2 r$ . Scaling: $m, v, r$ each up 20% ⇒ $F \times \dfrac{1.2 \times 1.44}{1.2} = 1.44F$ .
Same force, same radius, masses $m$ and $3m$ : speeds in ratio $\sqrt{3} : 1$ .
Spring of natural length $L$ : $kx = m\omega^2(L + x)$ .
Mass $m$ circling on a table, string through a hole to a hanging $M$ : $Mg = m\omega^2 L$ , $f = \dfrac{1}{2\pi}\sqrt{\dfrac{Mg}{mL}}$ .
String of length $l$ swept round a vertical axis (conical pendulum): the tension alone gives $T = ml\omega^2$ , whatever the angle.

Centripetal force

F = \frac{mv^2}{r} = m\omega^2 r

Worked example

A 0.5 kg mass moves at 4 m/s on a 2 m circle. Centripetal force? If the mass is doubled, the speed halved and the radius halved, what is it then?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q11Moderate

Example 1 · Rotational Dynamics · Dynamics of Circular Motion — Banking, Conical Pendulum, Vertical Circle

A particle of mass 'm' performs uniform circular motion of radius 'r' with linear speed 'v' under the application of force 'F'. If 'm', 'v' and 'r' are all increased by 20%, the necessary change in force required to maintain the particle in uniform circular motion is

Adding percentages

Scaling is multiplicative:

1.2 \times 1.2^2 \div 1.2 = 1.44

, a 44% rise. Adding the three 20%s is how 12% and 14% get into the options.

Drill 5 more on the centripetal force and what supplies it

Concept 2 of 3: Forces at an Angle: Banked Roads, Conical Pendulums and Funnels

Whenever a single force is tilted — the normal reaction of a banked road, the string of a conical pendulum, a funnel wall — its vertical part holds the weight and its horizontal part turns the body. Divide one by the other and tan θ = v²/rg falls out every time.

Definition

Banked road with no friction needed: $\tan\theta = \dfrac{v^2}{rg}$ . Outer edge raised $h$ over width $b$ : $h = \dfrac{v^2b}{rg}$ .
Same banking and friction: $v_{\max} \propto \sqrt{r}$ ; 20% more speed needs 44% more radius.
Bob hanging in a car rounding a curve: string at $\tan^{-1}\dfrac{v^2}{rg}$ to the vertical.
Conical pendulum of length $l$ at angle $\theta$ : $\omega = \sqrt{\dfrac{g}{l\cos\theta}}$ .
Smooth funnel: the circle of speed $V$ sits a height $h = \dfrac{V^2}{g}$ above the vertex.
Humped (convex) road: $N = mg - \dfrac{mv^2}{r}$ ; dipped (concave): $N = mg + \dfrac{mv^2}{r}$ , the largest.

A tilted force turning a body

\tan\theta = \frac{v^2}{rg}

Worked example

A road of radius 50 m is to be banked for 10 m/s. Banking angle, and how high the outer edge of an 8 m wide road must be?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q29Moderate

Example 2 · Rotational Dynamics · Dynamics of Circular Motion — Banking, Conical Pendulum, Vertical Circle

Radius of curved road is '

R

', width of road is '

b

'. The outer edge of road is raised by '

h

' with respect to inner edge so that a car with velocity '

V

' can pass safe over it, then value of '

h

' is (

g =

acceleration due to gravity)

Asking for the new radius, answering the increase

'The increase in the radius of curvature' for 20% more speed on a 20 m curve is 8.8 m; the new radius, 28.8 m, sits beside it in the options.

Drill 5 more on forces at an angle: banked roads, conical pendulums and funnels

Concept 3 of 3: The Vertical Circle

In a vertical circle gravity helps at the top and fights at the bottom, so the tension is least at the top and greatest at the bottom, and energy conservation links the speeds: v²(bottom) = v²(top) + 4gr. To just complete the loop the string must just stay taut at the top.

Definition

Top: $T_{\text{top}} = \dfrac{mv_t^2}{r} - mg$ ; bottom: $T_{\text{bot}} = \dfrac{mv_b^2}{r} + mg$ ; $v_b^2 = v_t^2 + 4gr$ .
So $T_{\text{bot}} - T_{\text{top}} = 6mg$ , always.
Just completing the loop: $v_t = \sqrt{gr}$ , $v_b = \sqrt{5gr}$ , and the apparent weight at the bottom is $6mg$ .
A thread that bears $T_{\max}$ : the stone's speed is limited at the BOTTOM, $T_{\max} = mg + m\omega^2 r$ .
With gravity the only force doing work, the total mechanical energy is the same at every point.

Vertical circle

T_{\text{bottom}} - T_{\text{top}} = 6mg, \qquad v_{\text{top,min}} = \sqrt{gr}, \quad v_{\text{bottom,min}} = \sqrt{5gr}

Worked example

A stone on a 0.4 m string is whirled in a vertical circle. Least speeds at the top and at the bottom for it to complete the circle? (g = 10)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q23Moderate

Example 3 · Rotational Dynamics · Dynamics of Circular Motion — Banking, Conical Pendulum, Vertical Circle

A body of mass '

m

' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is (

g =

gravitational acceleration)

Taking the bottom tension as 5mg

At the bottom

T = mg + \frac{mv^2}{r}

and

v^2 = 5gr

when just looping, so

T = 6mg

. The

5mg

forgets the weight.

Drill 4 more on the vertical circle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Centripetal Force and What Supplies It
Centripetal force
$F = \frac{mv^2}{r} = m\omega^2 r$
Forces at an Angle: Banked Roads, Conical Pendulums and Funnels
A tilted force turning a body
$\tan\theta = \frac{v^2}{rg}$
The Vertical Circle
Vertical circle
$T_{\text{bottom}} - T_{\text{top}} = 6mg, \qquad v_{\text{top,min}} = \sqrt{gr}, \quad v_{\text{bottom,min}} = \sqrt{5gr}$

Watch out for (3)

Adding percentages
→ The Centripetal Force and What Supplies It
Asking for the new radius, answering the increase
→ Forces at an Angle: Banked Roads, Conical Pendulums and Funnels
Taking the bottom tension as 5mg
→ The Vertical Circle

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