Q.A stone tied to the end of a string 80cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25s, what is the magnitude and direction of acceleration of the stone?
Uniform circular motion is motion along a circular path at constant speed. Although the speed stays the same, the velocity does not — its direction keeps changing at every instant — so the motion has an acceleration even though the speed never changes. This concept covers the kinematics of that motion; the force that causes it (centripetal force) belongs to the Laws of Motion unit.
1. Angular Quantities
As the body sweeps through an angle θ, its angular velocity is
ω=dtdθ
For one full revolution in a period T, at frequency f:
ω=T2π=2πf,T=f1
To convert rpm to rad/s:
ω=602π⋅(rpm)=30π(rpm)
The linear (rim) speed v and the angular speed ω are related by
v=ωr
2. Centripetal Acceleration
Even though the speed is constant, the velocity vector keeps turning — this produces an acceleration directed toward the centre of the circle:
ac=rv2=ω2r=T24π2r=4π2f2r
Use whichever form matches the data you are given. This acceleration is often expressed as a multiple of g (as ac/g, taking g=10m/s2).
3. Directions
The velocity is always tangential (along the direction of motion); the acceleration is centripetal — radial, pointing inward, and perpendicular to the velocity.
Because the direction of the velocity keeps changing, the change in the velocity vector over an angle Δθ has magnitude
∣Δv∣=2vsin(2Δθ)
So a quarter turn gives ∣Δv∣=v2, a half turn gives 2v, and a full turn gives 0.
The average acceleration over an arc is ∣Δv∣/Δt — this is smaller in magnitude than the instantaneous centripetal acceleration v2/r, and its direction lies along the perpendicular bisector of the chord joining the two points (which, for a circle, always passes through the centre) rather than being radially inward from either endpoint's own position.
4. Points on a Rotating Body
Every point on a rigid rotating body (a wheel, a disc, a clock hand, the Earth) shares the same ω, but the rim speed v=ωr grows with the radius.
Clock hands: the second hand turns at ω=602π rad/s, the minute hand at 36002π rad/s, the hour hand at 432002π rad/s.
The Earth spins with ω=864002π≈7.3×10−5 rad/s. A point on the equator moves at ωR≈465 m/s, and a point at latitude λ moves at ωRcosλ (since the circle of latitude has radius Rcosλ).
5. Connected Systems
Wheels joined by a belt (or two gears in mesh) share the same rim speed, so ω1r1=ω2r2 — the larger wheel turns with the smaller ω.
Wheels on the same axle (concentric) share the same ω, so the outer rim moves faster (v∝r).
6. Non-Uniform Circular Motion
If the speed also changes, there is a tangential acceleration
at=dtdv=rα
(from the angular acceleration α=dω/dt), in addition to the radial ac=v2/r. These two are perpendicular, so the total acceleration is
Concept: Centripetal acceleration in uniform circular motion. A stone in uniform circular motion accelerates toward the centre even at constant speed, because its direction of motion keeps changing.
With the string as the radius (0.8m) and the given rate of revolution, the stone's centripetal acceleration works out to about 9.9m/s2, directed radially inward toward the centre of the circle at every instant.
Why the stone accelerates at all
Even though the stone moves at constant speed, it is still accelerating — because velocity is a vector, and its direction keeps changing as the stone goes around the circle. This change in direction is exactly what centripetal acceleration describes: an acceleration that always points toward the centre, continuously turning the velocity vector without changing its magnitude.
The magnitude of centripetal acceleration can be written as:
ac=ω2r=rv2
Since we're given the number of revolutions and the time, the angular-velocity route is the most direct.
Step 1 — Radius
The string length is the radius of the circular path:
r=80cm=0.8m
Step 2 — Angular velocity
The stone makes 14 revolutions in 25s. Each revolution is 2π radians, so:
Concept: Centripetal Acceleration from the Linear Speed, Not the Angular Speed
Method: ac=v2/r via Distance Travelled Along the Circle (Not ac=ω2r)
Both existing solutions compute the angular velocity ω first and use ac=ω2r. This method never computes ω at all — it finds the stone's actual linear speed v (distance travelled around the circle, divided by time) and substitutes directly into the equally-valid form ac=v2/r.
Step 1 — Radius
r=80cm=0.8m
Step 2 — Total distance travelled, via the circle's circumference
Each revolution covers a distance equal to the circle's circumference, 2πr:
circumference=2π(0.8)=1.6π≈5.027m
In 14 revolutions, the stone covers:
d=14×5.027≈70.37m
Step 3 — Linear speed
v=td=25s70.37m≈2.815m/s
Step 4 — Centripetal acceleration, directly from v and r
ac=rv2=0.8(2.815)2=0.87.924≈9.9m/s2
Step 5 — Direction
As always in uniform circular motion, this acceleration points radially inward, toward the centre of the circular path traced by the string — this part of the reasoning doesn't depend on which formula was used to get the magnitude.
Why the linear-speed route is a genuine alternative …