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Asked by Chinmayee Last Modified
As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.
Firstly, let's break down the problem. We have a stone tied to the end of a string and whirled in a horizontal circle with constant speed. We're given that the string is 80 cm long and the stone makes 14 revolutions in 25 seconds. We need to find the magnitude and direction of acceleration.
To find the magnitude of acceleration, we can use the formula for centripetal acceleration:
ac=v2rac=rv2
Where:
To find vv, we'll use the formula for speed:
v=2πrTv=T2πr
Where:
Given that the stone makes 14 revolutions in 25 seconds, we can find TT by dividing the total time by the number of revolutions:
T=25 seconds14 revolutionsT=14 revolutions25 seconds
Now, let's calculate TT:
T=2514≈1.79 seconds per revolutionT=1425≈1.79 seconds per revolution
Now, we'll find vv:
v=2π×80 cm1.79 s≈282 cm/sv=1.79 s2π×80 cm≈282 cm/s
Now, let's plug vv and rr into the centripetal acceleration formula:
ac=(282 cm/s)280 cm≈997 cm/s2ac=80 cm(282 cm/s)2≈997 cm/s2
So, the magnitude of the acceleration of the stone is approximately 997 cm/s2997 cm/s2.
As for the direction of the acceleration, it's always directed towards the center of the circular path. This is because centripetal acceleration is the acceleration required to keep an object moving in a circle, and it acts perpendicular to the velocity of the object, towards the center of the circle.
Thus, the direction of the acceleration of the stone is towards the center of the circular path.
In summary, the magnitude of the acceleration of the stone is approximately 997 cm/s2997 cm/s2, and its direction is towards the center of the circular path.
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