A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s what is the acceleration of the stone?

Question

A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s what is the acceleration of the stone? (A) ((88/25))2 m s -2 (B) ((25/88))2 m s -2 (C) ((88/25)) m s -2 (D) ((25/88)) m s -2

Tardigrade Admin · Accepted Answer

Here, r=100 cm =1 m , v =(14/25) s -1 ∴ ω=2 π v=2 × (22/7) × (14/25)=(88/25) rad s -2 Centripetal acceleration, ac=ω2 r=((88/25))2 × 1.0 =((88/25))2 m s -2

Q. A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s what is the acceleration of the stone?

$(\frac{88}{25})^{2} m s^{- 2}$

$(\frac{25}{88})^{2} m s^{- 2}$

$(\frac{88}{25}) m s^{- 2}$

$(\frac{25}{88}) m s^{- 2}$

Here, $r = 100 c m = 1 m, v = \frac{14}{25} s^{- 1}$
$∴ ω = 2 π v = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} r a d s^{- 2}$
Centripetal acceleration,
$a_{c} = ω^{2} r = (\frac{88}{25})^{2} \times 1.0$
$= (\frac{88}{25})^{2} m s^{- 2}$

Q. A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s what is the acceleration of the stone?

(2588​)2ms−2

(8825​)2ms−2

(2588​)ms−2

(8825​)ms−2

Here, r=100cm=1m,v=2514​s−1 ∴ω=2πv=2×722​×2514​=2588​rads−2 Centripetal acceleration, ac​=ω2r=(2588​)2×1.0 =(2588​)2ms−2

$(\frac{88}{25})^{2} m s^{- 2}$

$(\frac{25}{88})^{2} m s^{- 2}$

$(\frac{88}{25}) m s^{- 2}$

$(\frac{25}{88}) m s^{- 2}$

Here, $r = 100 c m = 1 m, v = \frac{14}{25} s^{- 1}$
$∴ ω = 2 π v = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} r a d s^{- 2}$
Centripetal acceleration,
$a_{c} = ω^{2} r = (\frac{88}{25})^{2} \times 1.0$
$= (\frac{88}{25})^{2} m s^{- 2}$