A bob attached to a string oscillates back and forth like a simple pendulum. Resolve the forces acting on the bob into components. What is the acceleration experienced by the bob when the string makes an angle $\theta$ with the vertical?

A bob attached to a string oscillates back and forth like a simple pendulum. Resolve the forces acting on the

V. Numerical Problems · Q5

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Concept understanding — Centripetal Force

Centripetal Force: The Invisible Hand That Keeps Things Going in Circles

Imagine you're in a car taking a sharp turn to the left. You feel yourself being pushed to the right, against the door. That feeling — that's your body trying to keep moving straight while the car turns. Now here's the key insight: you don't actually feel a force pushing you outward. What you feel is your own inertia — your body's natural desire to keep moving in a straight line.

The real force is the one the car door exerts on you, pushing you inward toward the centre of the turn. That inward push is centripetal force.

The Intuition: Why Does Anything Need a Force to Go in a Circle?

Newton's first law says: an object in motion stays in motion in a straight line unless acted on by an external force. A straight line is the "default" path. To make something go in a circle — which is a constantly changing direction — you need a force that continuously pulls it away from that straight line.

Think of a stone tied to a string, whirled around your head. The string is taut. That tension is the centripetal force. If you let go, the stone doesn't fly outward — it flies off tangentially, in a straight line from the point of release. The string was constantly pulling it inward, preventing it from escaping.

Note

The word "centripetal" comes from Latin: centrum (centre) + petere (to seek). It means "centre-seeking." This is the opposite of "centrifugal" (centre-fleeing), which is a fictitious force you feel only in a rotating reference frame — not a real force in physics.

The Precise Statement

Centripetal force is any force that causes an object to follow a curved path, directed toward the centre of curvature of that path. It is not a new, independent force like gravity or friction. It is the name we give to the net force that points radially inward when an object moves in a circle.

For uniform circular motion (constant speed $v$ along a circle of radius $r$ ), the magnitude of centripetal force is:

F_c = \frac{m v^2}{r}

Where:

$m$ = mass of the object
$v$ = speed (magnitude of velocity)
$r$ = radius of the circular path

The corresponding centripetal acceleration (which is always perpendicular to velocity) is:

a_c = \frac{v^2}{r}

This acceleration points toward the centre. It is not constant in direction — it rotates as the object moves — but its magnitude is constant for uniform circular motion.

What Provides the Centripetal Force?

Centripetal force is always supplied by some real physical interaction. Here are common examples:

Situation	What provides centripetal force
Car turning on a flat road	Friction between tyres and road
Satellite orbiting Earth	Gravitational attraction
Stone on a string	Tension in the string
Electron orbiting a nucleus	Electrostatic attraction
A roller coaster looping the loop	Normal force from the track (plus gravity at the top)

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Step 1. A bob of mass m, on a string of length L, swings through an angle θ from the vertical. Two forces act on it: gravity mg (straight down) and string tension T (along the string, toward the pivot).

Step 2. Resolve gravity into two components relative to the string's direction: $mg\cos\theta$ (along the string, radially outward, opposing T) and $mg\sin\theta$ (perpendicular to the string, tangential to the bob's arc).

Step 3. TANGENTIAL direction: the only force component here is $mg\sin\theta$ (T has no tangential component). So the tangential acceleration is $a_t=g\sin\theta$ — this speeds the bob up or slows it down as it swings.

Step 4. RADIAL (centripetal) direction: the net inward force is $T-mg\cos\theta$ , so the centripetal acceleration is $a_c=\dfrac{T-mg\cos\theta}{m}$ , directed toward the pivot. …

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Q.A bob attached to a string oscillates back and forth like a simple pendulum. Resolve the forces acting on the bob into components. What is the acceleration experienced by the bob when the string makes an angle θ\thetaθ with the vertical?

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Q.A bob attached to a string oscillates back and forth like a simple pendulum. Resolve the forces acting on the bob into components. What is the acceleration experienced by the bob when the string makes an angle $\theta$ with the vertical?

Unlock everything free for 14 days