Kinematics Question 802
Question: A stone ties to the end of a string $ 1m $ 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 [CBSE PMT 2005]
Options:
A) $ \frac{{{\pi }^{2}}}{4}m{{s}^{-2}} $ and direction along the radius towards the centre
B) $ {{\pi }^{2}}m{{s}^{-2}} $ and direction along the radius away from the centre
C) $ {{\pi }^{2}}m{{s}^{-2}} $ and direction along the radius towards the centre
D) $ {{\pi }^{2}}m{{s}^{-2}} $ and direction along the tangent to the circle
Show Answer
Answer:
Correct Answer: C
Solution:
a $ =\frac{v^{2}}{r}={{\omega }^{2}}r $
$ =4{{\pi }^{2}}n^{2}r=4{{\pi }^{2}}{{( \frac{22}{44} )}^{2}}\times 1 $
$ ={{\pi }^{2}}m/s^{2} $ and it’s direction is always along the radius and towards the centre.
