A ball on the end of a string is whirled around
in a horizontal circle of radius 0.486 m. The
plane of the circle is 1.83 m above the ground.
The string breaks and the ball lands 2.39 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.The acceleration of gravity is 9.8 m/s^2
Find the centripetal acceleration of the ball
during its circular motion.?
To find the centripetal acceleration of the ball, we need to use the formula:
a = v^2 / r
Where:
- a is the centripetal acceleration
- v is the velocity of the ball
- r is the radius of the circle
From the given information, we can find the velocity of the ball right before the string breaks. We know that the ball is whirling in a horizontal circle, so its initial velocity is horizontal.
First, we need to find the time it takes for the ball to hit the ground after the string breaks. We can use the equation of motion in the vertical direction:
y = y0 + v0y * t - 0.5 * g * t^2,
Where:
- y is the distance the ball lands from the point on the ground directly beneath it (2.39 m)
- y0 is the initial height of the plane (1.83 m)
- v0y is the initial vertical velocity (0 m/s, as the ball is moving in a horizontal circle)
- g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the values:
2.39 = 1.83 + 0 * t - (0.5 * 9.8 * t^2)
Rearranging the equation:
4.9 * t^2 = 0.56
t^2 = 0.56 / 4.9
t^2 = 0.1143
t = sqrt(0.1143)
t ≈ 0.338
Since the ball is traveling in a horizontal circle, its horizontal velocity remains constant. The distance traveled along the horizontal direction is the circumference of the circle, given by:
C = 2 * π * r
Plugging in the values:
C = 2 * π * 0.486
C ≈ 3.05
The time taken to complete one circle is the same as the time it takes for the ball to hit the ground. Therefore, we can calculate the horizontal velocity of the ball before the string breaks:
v = C / t
v ≈ 3.05 / 0.338
v ≈ 9.04 m/s
Now, we can find the centripetal acceleration using the formula:
a = v^2 / r
a = (9.04)^2 / 0.486
a ≈ 168.16 m/s^2
Therefore, the centripetal acceleration of the ball during its circular motion is approximately 168.16 m/s^2.
To find the centripetal acceleration of the ball during its circular motion, you can use the equation:
Centripetal acceleration (ac) = (Velocity squared (v^2)) / Radius (r)
To solve this, we need to find the velocity of the ball when the string breaks.
First, let's find the time it takes for the ball to fall to the ground.
We can use the equation of motion:
Displacement (d) = Initial velocity (u) * time (t) + (1/2) * Acceleration (a) * time (t)^2
Since the ball starts from rest (u = 0), the equation simplifies to:
Displacement (d) = (1/2) * Acceleration (a) * time (t)^2
Plugging in the given values:
d = 1.83 m (height of the plane above the ground)
a = 9.8 m/s^2 (acceleration due to gravity)
1.83 m = (1/2) * 9.8 m/s^2 * t^2
Solving for time (t):
t^2 = (2 * 1.83 m) / 9.8 m/s^2
t^2 = 0.374 m / 9.8 m/s^2
t^2 = 0.038 m/s^2
t = √0.038 m/s^2
t = 0.195 s (approximated to three decimal places)
Now let's find the horizontal component of the initial velocity of the ball (v).
We can use the equation:
Horizontal displacement (dx) = Horizontal velocity (vx) * time (t)
Since the ball lands 2.39 m away from the point on the ground directly beneath it:
dx = 2.39 m
t = 0.195 s (from previous calculation)
2.39 m = (vx) * 0.195 s
Solving for horizontal velocity (vx):
vx = 2.39 m / 0.195 s
vx = 12.256 m/s (approximated to three decimal places)
Finally, to find the centripetal acceleration, use the formula:
Centripetal acceleration (ac) = (vx^2) / r
Plugging in the values:
ac = (12.256 m/s)^2 / 0.486 m
ac = 154.247 m^2/s^2 (approximated to three decimal places)
Therefore, the centripetal acceleration of the ball during its circular motion is approximately 154.247 m^2/s^2.