Question

a particle moves in a circle of radius 20 cm at 120 revolutions per minute .calculate the acceleration of particle​

Answer 1

Answer

• Acceleration of the particle = 4 π² m/s²

Given

• A particle moves in a circle of radius 20 cm at 120 revolutions per minute

To Find

• Acceleration of particle , a

Formula

$\bullet \;\;\; \rm \omega=2 \pi f\\\\\bullet \;\;\; \rm a=r \omega ^2$

Solution

Radius , r = 20 cm = 0.2 m

Frequency , f = 120 rev / m

⇒ f = 120 rev / 60 s

⇒ f = 2 rev/s

Now , Angular velocity ,

$\rm \omega = 2 \pi f\\\\\implies \rm \omega=2\pi (2)\\\\\implies \rm \omega=4 \pi \; rad/s$

Acceleration of the particle ,

$\rm a=r \omega ^2\\\\\implies \rm a=0.2 \times (4 \pi )^2\\\\\implies \rm a=0.25 \times 16\pi^2\\\\\implies \rm a=4\pi^2 \; m/s^2\\\\\implies \rm a=39.438\ m/s^2$

Answer 2

Radius (r) = 20 cm = 0.2 m

Frequency (f) = 120 rev / m

f = 120 rev / 60s

f = 2 rev/s

Now , Angular velocity ,

ω = 2πf

ω = 2π(2)

ω = 4πrad/s

Acceleration of the particle ,

a = 0.2×(4π)²

a = 0.25×16π²

a = 4π² m/s²

a = 39.438 m/s²

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