Search any question & find its solution
Question:
Answered & Verified by Expert
A stone tied to the end of a string $80 \mathrm{~cm}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in $25 \mathrm{~s}$ what is the magnitude and direction of acceleration of the stone?
Solution:
2498 Upvotes
Verified Answer
Given, $r=80 \mathrm{~cm}=0.8 \mathrm{~m}, n=14 / 25 \mathrm{~s}^{-1}$,
$$
\omega=2 \pi v=2 \times 22 / 7 \times 14 / 25=88 / 25 \mathrm{rad}^{-1} \mathrm{~s}^{-1}
$$
Centripetal acceleration, $a=v^2 / r=(250)^2 / 1000$
Again, $a / g=(250)^2 / 1000 \times 1 / 9.8=6.38$
$$
\omega=2 \pi v=2 \times 22 / 7 \times 14 / 25=88 / 25 \mathrm{rad}^{-1} \mathrm{~s}^{-1}
$$
Centripetal acceleration, $a=v^2 / r=(250)^2 / 1000$
Again, $a / g=(250)^2 / 1000 \times 1 / 9.8=6.38$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.