Astone at the end of a sling is whirled in a vertical circle of radius 1.10maeac

Question

Astone at the end of a sling is whirled in a vertical circle of radius 1.10maeaconstant speed Vo=1.00 m/s as in Figure P4.57 The center of the string is 1.50 m above the ground. Figure P4.57 ) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at A? (b) What is the range of the stone if it is released when the sling is inclined at 30.0 with the horizontal at B? (c) What is the acceleration of the stone just before it is released at A? Magnitude (absolute value) Direction O downward Oupward toward the center of the circle O in the direction of vo (d) What is the acceleration of the stone just after it is released at A? Magnitude (absolute value) Direction O toward the center of the circle O upward O downward O in the direction of vo

Explanation / Answer

Let’s determine the vertical and horizontal component of the stone’s initial velocity at both positions.


At A:
Vertical = 1 * sin 30 = 0.5 m/s
Horizontal = 1 * cos 30 = 0.866 m/s

The next step is to determine the vertical distance from point A to the ground. Let’s use the following equation to determine the vertical distance from point A to the dashed horizontal line.

d = r * sin = 1.10 * sin 30 = 0.55 meter
To determine the vertical distance from point A to the ground add 1.50 meters.
d = 2.05 meters
This is true for points A and B.

During the time the stone is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the time.

d = vi * t – ½ * g * t^2
d is the stone’s vertical displacement.
Vertical displacement = final height – initial height = 0 – 2.05 = -2.05 meters
g = 9.8 m/s^2

-2.05 = 0.5 * t – ½ * 9.8 * t^2

This is approximately 0.7 seconds. To determine the range, multiply the time by the stone’s initial horizontal velocity.

Range = 1 * cos 30 *0.7
This is approximately 0.61 meters.

For the stone at point B, the direction of its initial vertical velocity is downward. So its acceleration is +9.8 m/s^2. The total vertical distance it falls is 2.05 meters. Let’s use the following equation to determine the time.


d = vi * t + ½ * g * t^2
-2.05 = 0.5 * t + ½ * 9.8 * t^2

This is approximately 0.593 seconds.

Range = 1 * cos 30 *0.593
This is approximately 0.51 meters.


(c) What is the acceleration of the stone just before it is released at A? -9.8 m/s^2

Magnitude (absolute value)______ and Direction:(in the direction of v0;downward;toward the center of the circle;upward)


Vertical velocity = 0.5 m/s downward

Horizontal = 0.866 m/s

Magnitude = [(0.5^2 + (0.866 m/s)^2] = 0.999 = 1 m/s
Tan = -0.5 ÷ ( 0.866 m/s) = -30

(d) What is the acceleration of the stone just after it is released at A? +9.8 m/s^2

Magnitude (absolute value)_____ and DIrection: (downward;toward the center of the circle;in the direction of v0;upward)


This is 1 m/s at +30.

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