A girl whirls a stone in a horizontal circle 1.5 m above the ground by means of a string 141 cm long. The string breaks, and the stone flies of horizontally and strike the ground 13.6 m away. What is the centripetal acceleration of the stone while in circular motion?
To find the centripetal acceleration of the stone, we can use the formula:
a = (v^2) / r
where a is the centripetal acceleration, v is the velocity of the stone, and r is the radius of the circular motion.
First, let's find the velocity of the stone. Since the stone flies off horizontally and strikes the ground, we can assume it was moving horizontally when the string broke.
The horizontal distance the stone travels, d, is given as 13.6 m. The time it takes for the stone to travel this distance is the same as the time it would have taken to complete one full rotation. Therefore, we can use the formula:
v = d / t
where v is the velocity, d is the distance, and t is the time.
Since the stone is moving horizontally, the time it takes to travel the distance d is the same as the time for one full rotation, which is the period T.
Next, let's find the period T of the circular motion. The period is the time it takes for one complete rotation and can be calculated using the formula:
T = 2πr / v
where T is the period, r is the radius, and v is the velocity.
In this case, the radius of the circular motion is given as 1.5 m above the ground. However, the length of the string is given as 141 cm, which is 1.41 m. Since the stone is whirling 1.5 m above the ground, the radius r is equal to the length of the string plus 1.5 m.
Now we have all the information needed to find the centripetal acceleration using the formula:
a = (v^2) / r
Let's proceed with the calculations:
1. Calculate the velocity of the stone:
v = d / t
= 13.6 m / T
2. Calculate the period T of the circular motion:
T = 2πr / v
= 2π(1.41 m + 1.5 m) / (13.6 m / T)
= 2π(2.91 m) / (13.6 m / T)
= (18.24π/T) s
3. Calculate the velocity of the stone:
v = 13.6 m / T
4. Calculate the centripetal acceleration:
a = (v^2) / r
Note: the radius r is the length of the string plus the height of the circular motion.
Substituting the values into the formula:
a = ((13.6 m / T)^2) / (1.41 m + 1.5 m)
= (13.6 m)^2 / ((18.24π/T) s * 2.91 m)
= (184.96 m^2/s^2) / (18.24π/T)
= 184.96 m^2/(s^2*T) / (18.24π)
= (184.96 / (18.24π)) * (1 / T)
= (184.96 / (18.24π)) * (1 / (18.24π/T) s)
= (184.96 / (18.24π)) * (T / (18.24π) s)
Simplifying the expression:
a = 10.075 / T
Therefore, the centripetal acceleration of the stone while in circular motion is 10.075 / T m/s^2.
To find the centripetal acceleration of the stone, we will use the formula:
\(a_c = \frac{{v^2}}{{r}}\)
First, let's find the velocity of the stone while it is in circular motion. We know that the radius of the circular motion is 1.41 m and the stone is 1.5 m above the ground. We will use this information to calculate the length of the hypotenuse formed by the radius and the height of the stone:
\(Hypotenuse = \sqrt{{Height^2 + Radius^2}} = \sqrt{{1.5^2 + 1.41^2}}\)
We can now find the velocity by dividing the horizontal distance traveled by the time taken:
\(Velocity = \frac{{Distance}}{{Time}}\)
From the problem statement, we know that the stone strikes the ground 13.6 m away. However, we do not have the time taken. To calculate the time, we can use the equation of motion for horizontal motion:
\(Distance = \frac{1}{2} \cdot a \cdot t^2\)
Substituting the values, we get:
\(13.6 = \frac{1}{2} \cdot a \cdot t^2\)
Simplifying further, we get:
\(27.2 = a \cdot t^2\)
Now we can solve for time:
\(t = \sqrt{{\frac{{27.2}}{{a}}}}\)
Plugging this value of time into the equation for velocity, we can solve for velocity:
\(Velocity = \frac{{13.6}}{{\sqrt{{\frac{{27.2}}{{a}}}}}}\)
Finally, substituting the values of velocity and radius into the centripetal acceleration formula, we get:
\(a_c = \frac{{\left(\frac{{13.6}}{{\sqrt{{\frac{{27.2}}{{a}}}}}}\right)^2}}{{1.41}}\)
Simplifying this equation will give us the centripetal acceleration of the stone while in circular motion.