Question

14. A stone is tied to one end of string 50 cm long
and is whirled in a horizontal circle with constant
speed. If the stone makes 10 revolutions in 20 s,
then what is the magnitude of acceleration of the
stone :-
(1) 493 cm/s
(2) 720 cm/s
(3) 860 cm/s2
(4) 990 cm/s​

Answer 1

Given :

• Length of the string = 50 cm
• Number of revolution made by the stone in 20 sec = 10

To Find :

• The Magnitude of acceleration of the stone

Solution :

Angular velocity of the stone is given by ;

$\\ \star \: {\boxed{\purple{\sf{ \omega = 2\pi n}}}}$

Here ,

• ω is angular velocity
• n is number of revolutions made per sec

Now , We have ;

• Here while performing a horizontal circular motion the length of the string becomes the radius. So , r = 50 cm = 0.5 m
• We are given that number of revolution made by the stone in 20 sec as 10 applying unitary method here , number of revolutions made per sec is 1/2. So , n = 1/2

Now , By applying the formulae ;

$\\ : \implies \sf \omega = 2 \times 3.14 \times \frac{1}{2} \: {s}^{ - 1}$

$\\ : \implies \sf \omega = 1 \times 3.14 \: {s}^{ - 1}$

$\\ : \implies{\underline{\boxed{\red{\mathfrak{ \omega = 3.14 \: {s}^{ - 1} }}}}} \: \bigstar$

Now , Centripetal acceleration on stone is ;

$\\ \star \: {\boxed{\purple{\sf{a_c = { \omega}^{2} r }}}}$

Here ,

• $\sf{a_c}$ is centripetal acceleration
• ω is angular velocity
• r is radius

We have ,

• ω = 3.14 s⁻¹
• r = 0.5 m

Substituting the values ;

$\\ : \implies \sf \: a_c = {(3.14)}^{2} \times 0.5$

$\\ : \implies \sf \: a_c = 3.14 \: {s}^{ - 1} \times 3.14 \: {s}^{ - 1} \times 0.5 \: m$

$\\ : \implies \sf \: a_c \approx 4.93 \: m {s}^{ - 2}$

$\\ : \implies \sf \: a_c = 4.93 \times 100 \: cm {s}^{ - 2}$

$\\ : \implies{\underline{\boxed{{\pink{\mathfrak{a_c \approx493 \: cm {s}^{ - 2} }}}}}} \: \bigstar$

Hence ,

• The magnitude of acceleration on stone is 493 cm/s². So , Option(1) is the required answer.

Answer 2

✦Given:-

➨ Radius = 50 cm $\:$

➨ In 20 seconds the stone makes 10 revolutions.

➨ In 1 seconds the revolution =10/20=1/2

➨ N = 1/2 Hz

_______________________________________

✦To Find:-

➨ The magnitude of acceleration of the stone.

_______________________________________

✦Solution:-

➨ Now,

✪( Let us use π 3.14)✪

➨ Angular Velocity ( ω ) = 2πn

➨ ω = 2 × 3.14 × 50

➨ ω = 3.14

✪Radial Acceleration = ω²r✪

➨Radial Acceleration = (3.14)² × 50 cm/ s

➨Radial Acceleration = 3.14 × 3.14 × 50 cm/s

✪Radial Acceleration = 492.98 cm/s = 493 cm/s✪

✦Hence, the magnitude of acceleration of the stone is 493 cm/s

_______________________________________