A) \[\frac{\pi }{2}\] \[m{{s}^{-2}}\]and direction along the radius towards the centre
B) \[{{\pi }^{2}}\,m{{s}^{-2}}\]and direction along the radius away from 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
Correct Answer: C
Solution :
Since, speed is constant throughout the motion, so it is a uniform circular motion. Therefore, its radial acceleration \[a=r{{\omega }^{2}}\] \[=r{{\left( \frac{2\pi n}{t} \right)}^{2}}=r\times \frac{4{{\pi }^{2}}{{n}^{2}}}{{{t}^{2}}}\] \[=\frac{1\times 4\times {{\pi }^{2}}\times {{(22)}^{2}}}{{{(44)}^{2}}}\] \[={{\pi }^{2}}m/{{s}^{2}}\] This acceleration is directed along radius of circle.
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