Question

A stone, tied to the end of a string of length 50 cm, is whirled in a horizontal circle with a constant speed. If the stone makes 40 revolutions in 20 s, what is the magnitude and the direction of the acceleration of the stone?​

Answer 1

Given :-

Length of the string = 50 cm

Number of revolutions in 20 sec = 40

To Find :-

The magnitude and the direction of the acceleration of the stone.

Solution :-

We know that,

• w = Angular velocity
• r = Radius
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf Angular \ velocity=2 \pi n}}$

Given that,

Revolution (n) = 2

Substituting their values,

w = 2 × 3.14 × 2

w = 12.56

Using the formula,

$\underline{\boxed{\sf Radial \ acceleration=w^2 r }}$

Given that,

Angular velocity (w) = 12.65

Radius (r) = 50 cm = 0.5 m

Substituting their values,

a = (12.56)² × 0.5

a = 157.7536 × 0.5

a = 78.87 m/s²

Therefore, the magnitude and the direction of the acceleration of the stone is 78.87 m/s².

Answer 2

$\huge{\boxed{\rm{Question}}}$

A stone, tied to the end of a string of length 50 cm, is whirled in a horizontal circle with a constant speed. If the stone makes 40 revolutions in 20 s, what is the magnitude and the direction of the acceleration of the stone?

$\huge{\boxed{\rm{Answer}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• String's length = 50 cm.
• Revolution' number in 20 seconds = 40 revolutions.

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The magnitude and the direction of the acceleration of the stone.

${\bigstar}\large{\boxed{\sf{\pink{☞}}}}$

$\large\purple{\texttt{We write}}$

• $\large\green{\texttt{Radius as r}}$
• $\large\green{\texttt{Angular Velocity as w}}$
• $\large\green{\texttt{Acceleration as a}}$
• $\large\green{\texttt{n means revolution}}$

Some formulas that are used in this question →

◕ Angular velocity = 2πn

◕ Radial equation = w²r

$\huge{\boxed{\rm{Full \: solution}}}$

${\bigstar}\large{\boxed{\sf{\pink{Formula \: to \: find \: angular \: velocity}}}}$

$\large\orange{\texttt{2πn}}$

Revolution = 2 {Given}

$\large\purple{\texttt{Substituting the value we get}}$

✪ w = 2πn

✪ w = 2 × 3.14 × 2

✪ w = 4 × 3.14

✪ w = 12.56

${\bigstar}\large{\boxed{\sf{\pink{Formula \: to \: find \: radial \: equation}}}}$

$\large\orange{\texttt{w²r}}$

✪ w = 12.56 {Finded}

✪ r = 50 cm {Given} = 0.5 m

$\large\purple{\texttt{Substituting the value we get}}$

✪ a = (12.56)² × 0.5

✪ a = 12.56 × 12.56 × 0.5

✪ a = 157.7536 × 0.5

✪ a = 78.87 m/s²

So, the magnitude and the direction of the acceleration of the stone is 78.87 m/s².

$\large{\boxed{\sf{\red{78.87 m/s² \: is \: answer}}}}$

@Itzbeautyqueen23

Hope it's helpful to you all.

Thank you guys :]