Question

A large blade of a helicopter is rotating in a horizontal circle with constant angular velocity. Determine the ratio of acceleration of two particles A and B which are located at a distance of $3\text{cm}$ and $6\text{cm}$ respectively from the center of the circle.

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:6$

Answer 1

The correct option is B

$1:2$

Given data:

A large blade of a helicopter is rotating in a horizontal circle with constant angular velocity.

Two particles are located at a distance of $r_1=3cm$ and $r_2=6cm$ from the centre of the circle.

Centripetal acceleration:
We know that the acceleration of the particle in a circular path is given as $a=r{\omega }^2$, where $r=$radius of the circular path and $\omega =$ angular velocity.
Step 2 : Now, according to the question, we get
$$\begin{gathered} \frac{a_1}{a_2}=\frac{r_1{\omega }^2}{r_2{\omega }^2} \\ \Rightarrow \frac{a_1}{a_2}=\frac{3}{6} \\ \Rightarrow \frac{a_1}{a_2}=\frac{1}{2} \\ \Rightarrow a_1:a_2=1:2 \end{gathered}$$

Final Answer: Hence, option (b) is correct.