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

Step 1 :- Find angular frequency.
Given, length of the string $=80\text{cm}=0.8m$, it will be equal to the radius of the circular path i.e. r = 0.8 m.
Number of revolutions (n) = 14.
Time taken = 25 s
Frequency $\left(v\right)=\frac{\text{no. of revolutions}}{\text{time taken}}=\frac{14}{25}Hz$
As we know angular frequency is given by $\left(\omega \right)=2\pi v$.
$\Rightarrow \omega =2\times \frac{22}{7}\times \frac{14}{25}=\frac{88}{25}{\text{rad s}}^{-1}$

Step 2 :- Calculate centripetal acceleration.
Centripetal acceleration,
$a_c={\omega }^{2}r$
Substituting the value of $\omega \&r$, we get
$a_c=\frac{88}{25}\times \frac{88}{25}\times 0.8=9.91{\text{ms}}^{-2}$
The direction of centripetal acceleration is always directed along the radius towards the centre of circular path.

Final Answer: The magnitude of the acceleration is $9.91{\text{ms}}^{-2}$ and it is directed along the radius towards the centre of the circular path.