Upon arriving at a playground a child immediately runs to the swing set.
1. Since this question was ambiguous it has been withdrawn. No answer is required.
2. As the boy swings back and forth freely, what is his angular velocity about the swings pivot point when he is at his highest position behind the swing's pivot, at his lowest point directly under the pivot, and at his highest point in front of the pivot? For the case when he is directly under the pivot give a descriptive answer, not a number or formula.
Answer: At his highest position behind the swing's pivot his angular velocity is zero. At his highest point in front of the swing his angular velocity is also zero. When he is at his lowest point directly under the pivot his angular velocity has its maximum value (size) during the motion, but can be in one of two directions since he is swinging back and forth. If we look at the boy toward his right hand side, the angular velocity is counterclockwise when he is moving forward (using the right hand rule the angular velocity vector points toward us) and clockwise when he is moving backward (the angular velocity vector points away from us).
Why: As the boy moves forward and upward a component of the force due to gravity provides a torque that slows him down (an angular deceleration). Eventually he comes to a complete stop. His stop is brief, however, since the force of gravity continues to provide a torque which accelerates him back toward where he started. For a brief moment at the bottom of his motion there is no angular acceleration. Then the angular acceleration suddenly switches direction and begins to increase again, but in a direction opposite his motion. He decelerates and slows down as he moves upward.
3. While he is swinging the ropes supporting the swing suddenly break when he is at the highest point in his motion. Where will he land?
Answer: Directly beneath his position where the ropes broke.
Why: At the highest point in his motion the boys velocity is zero. With the ropes broken he is now simply a freely falling object and will fall straight down.
4. The ropes are actually more likely to break at the bottom of the swing's motion. Why?
Answer: At the bottom of the swing's motion the ropes must support the entire weight of the boy. The ropes must also supply the additional force necessary to accelerate him along a circular path.
Why: When the boy stops at the high point in his motion and begins to move back down he is essentially freely falling. If it were really a free fall he would go straight down, but there are ropes pulling him in the direction of the pivot. As he continues to fall the ropes support more and more of his weight. At the bottom the support force provided by the ropes and the downward force due to gravity are in opposite directions and the boy's full weight must be supported. In addition to supporting the boy's weight the ropes are providing the acceleration that causes him to move along the arc of a circle. You'll learn more about the forces associated with accelerating objects when we discuss centrifuges and roller coasters.
Having broken the swing the child moves on to the slide. This slide has a steep downward slope followed by a short horizontal flat section at the bottom and is very smooth (no friction).
5. Describe the boy's velocity at the top of the slide. How is his velocity changing in the middle of the slide? How is his velocity changing on the flat section at the bottom of the slide?
Answer: The boy begins at the top of the slide at rest. As he slides down the steep slope his velocity increases due to an acceleration pointing down the slope of the slide. At the bottom flat section of the slide the boy continues to move horizontally, but now with a constant speed.
Why: The boy accelerates down the steep slope of the slide. This acceleration is provided by the combination of the force of gravity pointing vertically downward and the support force of the slide pointing normal to the surface of the slide. At the bottom, flat section of the slide, the boy is no longer being accelerated since the force of gravity is exactly canceled by the now vertically upward support force of the slide, so he moves at a constant speed.
6. If the slide had a more gradual downward slope, but still started from the same height how would the boy's velocity differ at the bottom of the slide?
Answer: The boy's velocity at the bottom of the slide would be unchanged. He would move horizontally along the flat section away from the sloped section with exactly the same speed as on the steeper slide.
Why: Although the size of the acceleration along the slide is smaller, the time the boy spends on the slide will be longer resulting in the same velocity at the end. Another way of arriving at this answer is using the conservation of energy. In going down the slide the boy will be losing the same amount of potential energy independent of its slope since the slides are the same height. Since we are neglecting friction all of this potential energy will be converted into kinetic energy resulting in equal velocities at the bottom.
7. How would his velocity differ if he slid down the slide carrying a heavy rock?
Answer: The boy will have the same velocity at the bottom of the slide even if he is carrying a very heavy rock.
Why: The force that the boy experiences is due to the combined forces of gravity and the support force provided by the slide. The support force cancels the component of the gravitational force (the boy's weight) that is normal to the slide. This keeps the boy from penetrating the slide. The remaining force is along the slide and is a fraction of the boy's original weight. Since his weight is proportional to mass and force is equal to mass times acceleration the masses cancel and his acceleration is independent of mass, but it is only a fraction of the full acceleration due to gravity.
8. If he dropped the rock over the side of the slide when he was half way down, would they hit the ground at the same time (assume the flat section at the bottom of the slide is at ground level)? Why?
Answer: The rock will hit the ground first since it experiences a larger vertical component of acceleration.
Why: After the boy drops the rock it accelerates only in the vertical direction due to gravity. The boy, however, accelerates along the direction of the slide. The boy lands in front of the rock since the horizontal component of his acceleration causes the boy's horizontal component of velocity to continue increasing while the rocks horizontal component of velocity remains constant. The rock lands before the boy since the rock experiences the full acceleration due to gravity in the vertical direction while for the boy, the support force of the slide cancels some of this gravitational force.