Question

A stone tied to the end of a string of 1m long is whorled in a horizontal circle woth a constant speed.If the stone makes 22 revolution in 44 seconds what is the magnitude of the acceleration of the stone in terms of pi?​

Answer 1

Here is your answer my friend.

$a = r {w}^{2}$

$w = 2\pi \: v$

22 revolution = 44 sec.

1 revolution = 2 sec = t

$v = \frac{1}{t}$

$v = \frac{1}{2} \: \: and \: \: \: {v}^{2} = \frac{1}{4}$

$a = r {w}^{2} ......(1)$

${w}^{2} = 4 {\pi}^{2} \times {v}^{2}$

${w}^{2} = 4 {\pi}^{2} \times \frac{1}{4}$

$r = 1$

putting all the values in eq 1.

$a = 1 \times \frac{4 {\pi}^{2} }{4}$

$a = {\pi}^{2} \: \: \: \frac{m}{ {s}^{2} }$

Answer 2

Answer:

Acceleration of the stone is π² m/s² and direction along the radius towards the Centre.

Explanation:

r = 1 m

f = 22/44 = 1/2

ω = 2πf = 2π ×1/2 = π

$a_{c}$ = rω² = 1×(π)² = π² m/s²

$a_{c}$ = centripetal acceleration ⇒ along the radius towards the Centre.