Kinematics Question 655
Question: A stone tied to the end of a string of 1 m long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone?
Options:
A) $ {{\pi }^{2}}\text{m}{{\text{s}}^{-2}} $ and direction along the radius towards the center.
B) $ {{\pi }^{2}}\text{m}{{\text{s}}^{-2}} $ and direction along the radius away from the center.
C) $ {{\pi }^{2}}\text{m}{{\text{s}}^{-2}} $ and direction along the tangent to the circle.
D) $ {{\pi }^{2}}\text{/4m}{{\text{s}}^{-2}} $ and direction along the radius towards the center.
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ {{\text{a}} _{\text{r}}}-{{\omega }^{\text{2}}}\text{R} $ $ a _{r}={{( 2\pi 2 )}^{2}}R=4{{\pi }^{2}}2^{2}R $
$ \text{=}\text{4}{{\pi }^{2}}{{( \frac{22}{44} )}^{2}}( 1 )\text{}[ \because \text{v}=\frac{22}{44} ] $ .
$ {{\text{a}} _{\text{t}}}\text{=}\frac{\text{dv}}{\text{dt}}=0 $
$ a _{net}=a _{t}={{\pi }^{2}}m{{s}^{-2}} $ and direction along the radius towards the center.
