Question

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

Answer 1

Given :

Length of the string = 80cm = 0.8m.

Number of revolutions = 14

Time taken = 25s

To find :

The magnitude and direction of acceleration of the stone.

Solution :

firstly we have to find the angular frequency,

frequency is given by number of revolutions by time taken that is

frequency = number of revolutions/time taken = 14/15

we know,

Angular frequency, ω = 2πv

where, v denotes frequency

ω = 2 × 22/7 × 14/25 = 88/25 rad/s.

The acceleration of an object in uniform circular motion always directed towards the center of the circle is known as centripetal accerlation.

when a particle is moving along a circle of radius with a uniform speed then centripetal accerlation is

$\leadsto \sf{a_c = \omega^2 r}$

here,

• ω denotes angular frequency
• r denotes length

$\leadsto \sf{a_c = { \bigg(\dfrac{88}{25} \bigg) }^{2} \times 0.8}$

$\leadsto \sf{a_c = 9.91 \: {ms}^{ - 2} }$

thus, the direction of centripetal accerlation is always directed along the string, toward the centre, at all points.