A boy whirls a stone in a horizontal circle of radius 1.9 m and at height 2.1 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?
The centripetal acceleration was V^2/R, where R = 1.9 m.
You need to compute V, the speed when in circular motion. First compute the time T that it takes the stone to fall 1.9 m.
1.9 m = (1/2) g T^2
V = Vx, the horizontal component of velocity after the string breaks
V = (10 m)/T
98
To find the magnitude of the centripetal acceleration of the stone while in circular motion, we need to use the formula for centripetal acceleration.
The formula for centripetal acceleration is given by:
ac = (v^2) / r
Where:
ac = centripetal acceleration
v = velocity of the object in circular motion
r = radius of the circular path
From the given information:
- The radius of the circular path (r) is 1.9 m.
- The stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m.
- The height of the circle above ground level is 2.1 m.
To find the velocity (v) of the stone, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
u = initial velocity
a = acceleration (in this case, it is the gravitational acceleration)
s = distance traveled vertically (height of the circle above the ground)
Since the stone is whirled horizontally, the initial vertical velocity (u) is 0.
Plugging in the values, we get:
v^2 = 0 + 2 * (-9.8 m/s^2) * 2.1 m
v^2 = -41.16 m^2/s^2 (velocity squared)
To find the velocity (v), we take the square root of the velocity squared:
v = sqrt(-41.16) m/s
v = 6.42 m/s
Now, we can calculate the centripetal acceleration (ac) using the formula:
ac = (v^2) / r
ac = (6.42 m/s)^2 / 1.9 m
ac = 41.17 m^2/s^2
Therefore, the magnitude of the centripetal acceleration of the stone while in circular motion is approximately 41.17 m^2/s^2.