You are on a swing with a chain \(4.0 \mathrm{~m}\) long. If your maximum displacement from the vertical is \(35^{\circ},\) how fast will you be moving at the bottom of the arc?

Short Answer

Expert verified
Answer: \(v = \sqrt{2g \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ})))}\)

Step by step solution

01

Calculate height at the highest point

We will first find the height between the swing at its highest point and at the bottom of its arc. To do this, we can represent the 35° angle as a triangle, with base as the adjacent side, and height as the opposite side. Using trigonometry, we can determine height from the 4.0m chain: \(h = 4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ}))\) Calculate the value of h:
02

Calculate the potential energy at the highest point

To calculate the potential energy at the highest point, we can use the formula: \(PE = mgh\) where m is the mass (we can keep the mass general for now), g is the gravitational acceleration (approximately \(9.8\mathrm{~m/s^2}\)), and h is the height calculated in Step 1. \(PE = mg \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ})))\)
03

Calculate the kinetic energy at the bottom of the arc

At the bottom of the arc, the swing has a potential energy of 0, as it is considered at the reference or base level. All the potential energy is converted into kinetic energy. Using the conservation of energy principle, we can write: \(PE = KE\) where KE is the kinetic energy at the bottom of the arc. \(mg \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ})))= KE\)
04

Calculate the speed at the bottom of the arc

To find the speed, we can use the formula for kinetic energy: \(KE = \frac{1}{2}mv^2\) where v is the speed of the swing at the bottom of its arc. Now, we can equate the kinetic energy formula from Step 3: \(mg \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ}))) = \frac{1}{2}mv^2\) Since m is present in both sides of the equation, it cancels out: \(g \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ}))) = \frac{1}{2}v^2\) Now, we can re-arrange and solve for v: \(v = \sqrt{2g \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ})))}\) Calculate the value of v:
05

State the result

From the calculations above, we have determined the speed of the swing at the bottom of its arc. So the speed of the swing is: \(v = \sqrt{2g \cdot (4.0\mathrm{~m} \cdot (1 - \cos(35^{\circ})))}\)

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Key Concepts

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

Conservation of Energy
When we push a swing and release it, the swing moves back and forth in a graceful arc, an act we take for granted. But look closer, and you'll find an intricate dance of energy transformations governed by the conservation of energy principle. Imagine that as you sit still on the swing, hoisted up at an angle, you are a storage vessel brimming with potential energy. This energy is the product of your mass, gravitational force, and the height above the ground.

Once you release and begin to descend, this potential energy doesn't just disappear. Like a meticulous accountant, nature ensures that every joule is accounted for, transforming it into kinetic energy - the energy of motion. Therefore, at the bottom of the swing's arc, where you are at your lowest point, the potential energy is at its minimum, while your kinetic energy and speed are at their maximum. This seamless conversion back and forth without loss in an isolated system is the essence of the conservation of energy.
Kinetic Energy Calculation
Now let's talk about kinetic energy, which is essentially the energy an object possesses due to its motion. When you approach the bottom of the swing's arc, you're moving the fastest and, consequently, have the most kinetic energy. The calculation for kinetic energy is expressed with the equation \(KE = \frac{1}{2}mv^2\), where m is the mass of the swinging object (in this case, you), and v is the velocity or speed.

What's important to understand is that kinetic energy increases rapidly with speed—since speed is squared in the equation. This makes intuitive sense; as you pick up speed on a swing, you feel the wind stronger against your face, which indicates you're gaining kinetic energy at a quickening pace. In the swing physics problem, by equating this kinetic energy to the initial potential energy, we're able to determine the speed at the bottom of the arc!
Potential Energy Calculation
Unlike kinetic energy, which is zero at the swing's highest point, potential energy is at its peak. Potential energy is the energy held by an object because of its position relative to other objects. It's like stretching a rubber band—pulling it back gives it potential to snap forward. In the context of a swing, lifting it to the starting height injects it with potential energy, according to the formula \(PE = mgh\), where m is mass, g is the acceleration due to gravity (approximately \(9.8\mathrm{~m/s^2}\)), and h is the height above the reference point.

For the swing problem, by knowing the length of the chains and the angle of displacement, we can pinpoint how high above the lowest point the swing is. Through trigonometry, we can calculate this height and, as a result, the potential energy at the highest point. This potential energy converts into kinetic energy as you swing down.
Trigonometry in Physics
To solve our swing problem, we apply trigonometry, a branch of mathematics dealing with triangles' sides and angles. In physics, trigonometry comes in handy for determining distances or heights that are not readily measurable. For the swing issue, when at its highest point, the swing forms an invisible right triangle. The chain is the triangle's hypotenuse, and using the cosine for the angle we know, we determine the adjacent side - which represents how much of the full chain's length contributes to the vertical height.

The application of trigonometry continues by calculating the height difference, h, using the equation \(h = 4.0 \mathrm{~m} \cdot (1 - \cos(35^{\textdegree}))\). The result is crucial for finding the potential energy and, hence, the speed of the swing. By breaking down the motion in this way, trigonometry allows us to peer into the swing's swinging secrets - even when our feet can't touch the ground.

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Most popular questions from this chapter

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

Can the kinetic energy of an object be negative? Can the potential energy of an object be negative?

A 20.0 -kg child is on a swing attached to ropes that are \(L=1.50 \mathrm{~m}\) long. Take the zero of the gravitational potential energy to be at the position of the child when the ropes are horizontal. a) Determine the child's gravitational potential energy when the child is at the lowest point of the circular trajectory. b) Determine the child's gravitational potential energy when the ropes make an angle of \(45.0^{\circ}\) relative to the vertical. c) Based on these results, which position has the higher potential energy?

A cannonball of mass \(5.99 \mathrm{~kg}\) is shot from a cannon at an angle of \(50.21^{\circ}\) relative to the horizontal and with an initial speed of \(52.61 \mathrm{~m} / \mathrm{s}\). As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?

A 1.00 -kg block initially at rest at the top of a 4.00 -m incline with a slope of \(45.0^{\circ}\) begins to slide down the incline. The upper half of the incline is frictionless, while the lower half is rough, with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.300\). a) How fast is the block moving midway along the incline, before entering the rough section? b) How fast is the block moving at the bottom of the incline?

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