Question

2. A stone tied to the end of a string is whirled in a circular path of radius 50 cm with a constant speed. If the stone makes 200 revolutions in 5 minutes, find the linear speed of the stone?​

Answer 1

GiveN :-

• A stone tied to the end of a string is whirled in a circular path of radius 50 cm with a constant speed.
• The stone makes 200 revolutions in 5 minutes .

To FinD :-

• The linear speed of the stone .

SolutioN :-

Given that. , a stone is tied with a string of lenght 50cm .It makes 200 revolutions in a minute . We need to find the linear speed of the stone . First of all we know that liner speed is equal to product of angular velocity and radius . That is , $\sf v = r \omega$ .So we need to Find the angular velocity is $\sf 2\pi n$ .

$\red{\frak{We \ Know }}\begin{cases} \textsf{ Angular velocity as } \bf\omega \\\textsf{ Frequency as \textbf{ n}} . \\\textsf{ Linear Speed as \textbf{v}}.\\\textsf{ Radius as \textbf{r}}. \end{cases}$

$\underline{\red{\bigstar}\textsf{ Using the Formula we have :- }}$

$\sf\dashrightarrow \pink{ v = r \omega }\\\\\sf\dashrightarrow v = r \times ( 2 \pi n ) \\\\\sf\dashrightarrow v = 2 \times \dfrac{50m}{100} \times \dfrac{22}{7}\times \dfrac{200}{5\times 60s} \\\\\sf\dashrightarrow v = \dfrac{1}{2}\times 2 \times 3.14 \times \dfrac{200}{300} m/s \\\\\sf\dashrightarrow v = 3.14 \times \frac{2}{3} \\\\\sf\dashrightarrow \underset{\blue{\sf Required \ speed}}{\underbrace{\boxed{\pink{\frak{ Linear \ speed = 2.09 \ m/s }}}}}$