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

Question

A stone tied at the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14 s, what is the magnitude of acceleration of the stone? (A) 90 m / s 2 (B) 100 m / s 2 (C) 110 m / s 2 (D) 120 m / s 2

Tardigrade Admin · Accepted Answer

Length of the string, l=80 cm =0.8 m Number of revolutions =25, Time taken =14 s ∴ v =(25/14)=1.78 s -1 ∴ ω=2 π v=2 π × 1.78=11.18 rad / s a=ω2 l=(11.18)2 ×(0.8) =99.99 m / s 2 ≈ 100 m / s 2

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

$90 m / s^{2}$

$100 m / s^{2}$

$110 m / s^{2}$

$120 m / s^{2}$

Length of the string, $l = 80 c m = 0.8 m$
Number of revolutions $= 25,$ Time taken $= 14 s$
$∴ v = \frac{25}{14} = 1.78 s^{- 1}$
$∴ ω = 2 π v = 2 π \times 1.78 = 11.18 r a d / s$
$a = ω^{2} l = (11.18)^{2} \times (0.8)$
$= 99.99 m / s^{2} \approx 100 m / s^{2}$

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

90m/s2

100m/s2

110m/s2

120m/s2

Length of the string, l=80cm=0.8m Number of revolutions =25, Time taken =14s ∴v=1425​=1.78s−1 ∴ω=2πv=2π×1.78=11.18rad/s a=ω2l=(11.18)2×(0.8) =99.99m/s2≈100m/s2

$90 m / s^{2}$

$100 m / s^{2}$

$110 m / s^{2}$

$120 m / s^{2}$

Length of the string, $l = 80 c m = 0.8 m$
Number of revolutions $= 25,$ Time taken $= 14 s$
$∴ v = \frac{25}{14} = 1.78 s^{- 1}$
$∴ ω = 2 π v = 2 π \times 1.78 = 11.18 r a d / s$
$a = ω^{2} l = (11.18)^{2} \times (0.8)$
$= 99.99 m / s^{2} \approx 100 m / s^{2}$