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Exercises · 3.14

Q.A stone tied to the end of a string 80 cm80\ \text{cm} long is whirled in a horizontal circle with a constant speed. If the stone makes 1414 revolutions in 25 s25\ \text{s}, what is the magnitude and direction of acceleration of the stone?

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With the string as the radius (0.8 m0.8\ \text{m}) and the given rate of revolution, the stone's centripetal acceleration works out to about 9.9 m/s29.9\ \text{m/s}^2, 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=v2ra_c = \omega^2 r = \frac{v^2}{r}

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=80 cm=0.8 mr = 80\ \text{cm} = 0.8\ \text{m}

Step 2 — Angular velocity

The stone makes 1414 revolutions in 25 s25\ \text{s}. Each revolution is 2π2\pi radians, so:

ω=14×2π25=28π25 rad/s≈3.52 rad/s\omega = \frac{14 \times 2\pi}{25} = \frac{28\pi}{25}\ \text{rad/s} \approx 3.52\ \text{rad/s}

Step 3 — Centripetal acceleration

ac=ω2r=(28π25)2×0.8a_c = \omega^2 r = \left(\frac{28\pi}{25}\right)^2 \times 0.8

First, ω2=784π2625\omega^2 = \dfrac{784\pi^2}{625}. Using π2≈9.8696\pi^2 \approx 9.8696: …

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