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Question: A stone is moved in a horizontal circle of radius \[4m\] by means of a string at a height of \[20m\]...

A stone is moved in a horizontal circle of radius 4m4m by means of a string at a height of 20m20m above the ground. The string breaks and the particle' flies off horizontally, striking the ground 10m10m away. The centripetal acceleration during circular motion is
A) 6.25 ms26.25\text{ }m{{s}^{-2}}
B) 12.5 ms212.5\text{ }m{{s}^{-2}}
C) 18.75 ms218.75\text{ }m{{s}^{-2}}
D) 25 ms225\text{ }m{{s}^{-2}}

Explanation

Solution

We need time and the velocity to find the acceleration of the circular motion, the time will be calculated using the displacement kinematic formula and velocity by the formula of range of the flying object and the acceleration or the centripetal acceleration will be found by using the formula:
Acentripetal=v2r{{A}_{\text{centripetal}}}=\dfrac{{{v}^{2}}}{r}
The formula for the velocity, vv is:
v=Rtv=\dfrac{R}{t}
The formula for the time taken to fall after string breaks:
t=2Sgt=\sqrt{\dfrac{2S}{g}}
where vv is the velocity with which the object flies after the string breaks off. rr is the radius of the circular motion, RR is the range of the object it flies after string breaks, tt is the time taken to land, SS is the distance from the ground, and the gravity is taken g=10ms2g=10m{{s}^{-2}}.

Complete step by step solution:
To find the acceleration, we need two other objects vv and rr i.e. velocity and radius of the string and stone circular motion.
The height at which the stone is suspended from the ground is given as 20m20m.
The acceleration (due to gravity) of the stone falling after the string breaks is 10ms210m{{s}^{-2}}.
Hence, the time taken to reach the ground after breaking is:
t=2Sgt=\sqrt{\dfrac{2S}{g}}
Placing the values of the variables in the formula, we get:
t=2×2010\Rightarrow t=\sqrt{\dfrac{2\times 20}{10}}
t=2sec\Rightarrow t=2\sec
Now with the time known, let us find the velocity of the stone, when moving in a circular motion:
v=Rtv=\dfrac{R}{t}
Placing the values of the variables in the formula i.e. Range of the fall is 10m10m and time is 2sec2\sec , we get:
v=102v=\dfrac{10}{2}
v=5m/sec\Rightarrow v=5m/\sec
As we have found all the variables required to the find the centripetal acceleration we place the values of the variables in the formula and we get the answer as:
Acentripetal=522{{A}_{\text{centripetal}}}=\dfrac{{{5}^{2}}}{2}
Acentripetal=6.25 m/s2\Rightarrow {{A}_{\text{centripetal}}}=6.25\text{ }m/{{s}^{2}}
Therefore, the centripetal acceleration during circular motion is 6.25 m/s26.25\text{ }m/{{s}^{2}}.

Note:
Centripetal force and centrifugal force are both different forces although they are usually applied for the circular motion, but in centripetal force the force is acted on the axis of the circular motion and in centrifugal force the force is acted outside the circular motion hence, the formula for centripetal force is:
Acentripetal=v2r{{A}_{\text{centripetal}}}=\dfrac{{{v}^{2}}}{r}
And for centrifugal force is:
Acentripetal=v2r{{A}_{\text{centripetal}}}=-\dfrac{{{v}^{2}}}{r}
Apart from this we used the range formula because after the string breaks the stone acts as a projectile.