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Chapter 8: Problem 44

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by whirling it rap- idly in a horizontal circle and releasing it at the right moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being 20.0 m above the base of the cliff. The stone lands on the ground below the cliff at a point X. The horizontal distance of point X from the base of the cliff (directly beneath the point of release) is thirty times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone at the moment of release

Short Answer

Expert verified
The angular speed of the stone is approximately 14.85 rad/s.

Step by step solution

01

Identify the Given Information

Identify and record the known variables: - Height of the cliff above the ground: 20.0 m - Horizontal distance (range) of the landing point, X: 30 times the radius of the circle We want to find the angular speed of the stone at the moment of release.
02

Relate Horizontal Range with Circular Motion

Since the horizontal distance is given as 30 times the radius of the circle, we can express it as:\[ X = 30r \]where \( r \) is the radius of the circle in which the stone is whirled.
03

Use Projectile Motion to Find Time of Flight

Use the formula for time of flight in projectile motion, where the stone lands at a different level:\[ t = \sqrt{\frac{2h}{g}} \]Given, \( h = 20.0 \) m and \( g = 9.8 \) m/s虏, we can calculate:\[ t = \sqrt{\frac{2 \times 20.0}{9.8}} \approx 2.02 \text{ s} \]
04

Determine Horizontal Velocity

The horizontal velocity \( v_x \) is constant throughout the flight and is related to angular speed. First, express the horizontal range in terms of velocity:\[ X = v_x \cdot t \]Substitute \( X = 30r \) and \( t = 2.02 \) s:\[ 30r = v_x \cdot 2.02 \]Then, arrange to find \( v_x \):\[ v_x = \frac{30r}{2.02} \]
05

Relate Horizontal Velocity to Angular Speed

The horizontal velocity \( v_x \) is related to the angular speed \( \omega \) by the equation:\[ v_x = \omega r \]Substitute \( \omega r \) for \( v_x \) from the earlier step:\[ \omega r = \frac{30r}{2.02} \]Since \( r \) is common, it cancels out:\[ \omega = \frac{30}{2.02} \approx 14.85 \text{ rad/s} \]
06

Conclusion

The angular speed of the stone at the moment of release is approximately 14.85 rad/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Speed
Angular speed refers to how fast an object moves along a circular path. It's measured in radians per second (rad/s), giving us the rate of rotation. For example, when a stone is whirled in a slingshot, it travels in a circular trajectory. Upon release, this circular motion contributes to the stone's trajectory as it becomes projectile motion.
To calculate angular speed,
  • Use the formula: \( \omega = \frac{v_x}{r} \), where \( \omega \) is the angular speed, \( v_x \) is the horizontal velocity, and \( r \) is the radius of the circle.
Determining the stone's angular speed provides insights into its velocity just before being released from the slingshot. As shown in the exercise solution, the angular speed of the stone was found to be approximately 14.85 rad/s, indicating a fast rotational movement.
Projectile Motion
Projectile motion involves two types of movements: a horizontal motion at constant velocity and a vertical motion under constant acceleration due to gravity. Once released, the stone treated like a projectile follows a curved path due to these combined motions.
To analyze projectile motion, consider the following:
  • The vertical motion is influenced by gravity, which pulls the stone downward with an acceleration of approximately 9.8 m/s虏.
  • The horizontal motion remains constant because no horizontal forces act on the stone after release (ideal conditions).
In the exercise, the stone's angular velocity was related to its horizontal velocity when determining the landing point. The time of flight, calculated using \( t = \sqrt{\frac{2h}{g}} \), showed how long the stone would remain airborne before hitting the ground.
Horizontal Velocity
Horizontal velocity plays a crucial role in determining how far the stone travels. Since horizontal velocity is constant (ignoring air resistance), it directly affects the horizontal distance鈥攖he stone's range.
Key points on horizontal velocity:
  • It's derived from the combination of angular speed and radius in circular motion, \( v_x = \omega r \).
  • Once the stone is released, the initial horizontal velocity helps determine where it will land.
In this context, the relationship \( X = v_x \cdot t \) was used to find the horizontal distance covered, which equated to 30 times the radius. Understanding horizontal velocity allows us to predict and calculate projectile paths, ensuring accuracy in applications like the slingshot scenario.

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Most popular questions from this chapter

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