A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed- If the stone makes 14 revolutions in 25 s- what is the magnitude and direction of acceleration of the stone -

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Toppr Admin · Accepted Answer

Frequency f-1425HzAngular velocity -x3C9-2-x3C0-f-2-xD7-14-x3C0-25rad-s-3-52rad-sAcceleration is a-x3C9-2r-3-522-xD7-0-8-9-9m-s2The direction is-xA0-along the radius at every point towards the centre-

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Frequency $f = \frac{14}{25} H z$
Angular velocity $ω = 2 π f = \frac{2 \times 14 π}{25} r a d / s = 3.52 r a d / s$
Acceleration is $a = ω^{2} r = {3.52}^{2} \times 0.8 = 9.9 m / s^{2}$
The direction is along the radius at every point towards the centre.

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Frequency f=1425HzAngular velocity ω=2πf=2×14π25rad/s=3.52rad/sAcceleration is a=ω2r=3.522×0.8=9.9m/s2The direction is along the radius at every point towards the centre.

Frequency $f = \frac{14}{25} H z$
Angular velocity $ω = 2 π f = \frac{2 \times 14 π}{25} r a d / s = 3.52 r a d / s$
Acceleration is $a = ω^{2} r = {3.52}^{2} \times 0.8 = 9.9 m / s^{2}$
The direction is along the radius at every point towards the centre.