Question

A stone tied to the end of the string 80cm long is whirled in a horizontal circle 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

Answer:

Length of string = 80cm= 0.8m

Stone makes 14 revolution in 25 sec

therefore, Stone makes 14/25 revolution in 1 sec

frequency= 14/25 sec^-1

We know

Angular speed (ω) = 2π×frequency

=2π×14/25

=28π/25 rad/sec

Now acceleration(a) =ω²r

where r is the radius { length of the string }

Acceleration =0.8 × {28π/25}²

= 9.91 m/s²

Explanation:

Hope this helps

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.