Question

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

Answer 1

given : 14 rev. in 25 sec. i.e 25/14 rev. in 1 sec. which is freq.
acc. =r(angular vel.) ^2
and angular vel. = 2 ×3.14× freq .


direction of acc. is always towards center cuz centripetal force is acting

Answer 2

Answer:

Acceleration of the stone is 9.91 m/s² and direction along the radius towards the Centre.

Explanation:

r = 80 cm = 0.8 m

f = 14/25 = 0.56

ω = 2πf = 2π ×0.56 = 3.52

$a_{c}$ = rω² = 0.8×(3.52)² = 9.91 m/s²

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