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BIO Gymnastics. We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 \(\mathrm{kg}\) and diameter 1.0 \(\mathrm{m}\) . If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

Short Answer

Expert verified
The tumbler has a total kinetic energy of approximately 138.61 J, with 33.3% being rotational.

Step by step solution

01

Convert Rev/s to m/s

To find the linear velocity, we need to convert the revolutions per second to meters per second. Since 1 revolution is the same as the circumference of the cylinder, the circumference is given by \( \pi \times \text{diameter} = \pi \times 1.0 \, \text{m} = \pi \, \text{m} \). Thus 0.50 rev/s is equivalent to a linear velocity \( v \) of \( 0.50 \times \pi \approx 1.57 \, \text{m/s} \).
02

Calculate the Moment of Inertia

The moment of inertia \( I \) for a solid cylinder is given by \( I = \frac{1}{2} m r^2 \), where \( m = 75 \, \text{kg} \) and the radius \( r = 0.5 \, \text{m} \). So, \( I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \, \text{kg} \cdot \text{m}^2 \).
03

Calculate Linear Kinetic Energy

The linear kinetic energy \( KE_{\text{linear}} \) is given by \( \frac{1}{2} m v^2 \). Using the mass \( m = 75 \, \text{kg} \) and velocity \( v = 1.57 \, \text{m/s} \), the linear kinetic energy is \( \frac{1}{2} \times 75 \times (1.57)^2 \approx 92.43525 \, \text{J} \).
04

Calculate Rotational Kinetic Energy

The rotational kinetic energy \( KE_{\text{rotational}} \) is given by \( \frac{1}{2} I \omega^2 \). Use \( I = 9.375 \, \text{kg} \cdot \text{m}^2 \) and \( \omega = 0.50 \times 2 \pi = \pi \, \text{rad/s} \). Thus, \( KE_{\text{rotational}} = \frac{1}{2} \times 9.375 \times (\pi)^2 \approx 46.1701768 \, \text{J} \).
05

Calculate Total Kinetic Energy

The total kinetic energy \( KE_{\text{total}} \) is the sum of linear and rotational kinetic energy: \( KE_{\text{total}} = 92.43525 \, \text{J} + 46.1701768 \, \text{J} \approx 138.6054268 \, \text{J} \).
06

Determine Percentage of Rotational Energy

To find the percentage of the rotational energy, divide the rotational energy by the total energy and multiply by 100: \( \frac{46.1701768}{138.6054268} \times 100 \approx 33.3\% \).

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Key Concepts

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

Moment of Inertia
When we talk about moment of inertia, we're essentially discussing an object's reluctance to change its rotational state. Just like how mass measures how much an object resists changes in linear motion, moment of inertia does the same for rotation. For a gymnast modeled as a solid cylinder, the moment of inertia is calculated using a formula specific to cylinders. This formula is:
  • \( I = \frac{1}{2} m r^2 \)
where \(m\) is the mass of the cylinder (or gymnast in our case) and \(r\) is the radius. In the provided example, the gymnast's mass is 75 kg, and the radius, being half of the diameter, is 0.5 meters. Inserting these values gives us:
  • \( I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \text{ kg} \cdot \text{m}^2 \)
This value represents how much rotational motion the gymnast has "resisted." The larger this value, the harder it is to change the gymnast's spin. This concept is central in understanding the energy and dynamics involved when gymnasts perform their routines.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. It's a crucial aspect of kinetic energy, especially when analyzing objects that are in rotational motion, like a gymnast tumbling. The formula to determine the rotational kinetic energy is:
  • \( KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \)
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The angular velocity \( \omega \) can be found from the rotation speed, given in revolutions per second. In the given scenario, we have an angular velocity of \( \pi \) radians per second. Given the moment of inertia from earlier steps:
  • \( KE_{\text{rotational}} = \frac{1}{2} \times 9.375 \times (\pi)^2 \approx 46.17 \text{ J} \)
This energy shows how much energy is used to keep the gymnast spinning. Understanding the rotational kinetic energy can help in assessing performance and safety in activities involving spins or rotations.
Linear Velocity
Linear velocity refers to the rate of change of an object’s position along a path. For a rolling gymnast modeled as a cylinder, linear velocity can be derived from rotational motion. This means converting how many times the gymnast spins per second into how fast they move along the ground. Each revolution corresponds to a distance equal to the cylinder's circumference.
  • The circumference is \( \pi \times \text{diameter} = \pi \times 1.0 \text{ m} = \pi \text{ m} \).
  • Given 0.50 revolutions per second, converting this to meters per second results in a velocity of \( 0.50 \times \pi \approx 1.57 \text{ m/s} \).
This velocity is crucial because it provides insights into the gymnast's speed as they roll forward. Linear velocity, together with rotational speed, helps fully characterize the gymnast’s motion, combining both translational and rotational motion into one comprehensive picture. It’s important for planning and understanding movements dynamically, especially in sports and physics-related studies.

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Most popular questions from this chapter

Asteroid Collision! Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass \(M\) , for the day to become 25.0\(\%\) longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

A \(392-\mathrm{N}\) wheel comes off a moving truck and rolls with- out slipping along a highway. At the bottom of a hill it is rotating at 25.0 \(\mathrm{rad} / \mathrm{s} .\) The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J} .\) Calculate \(h .\)

A \(500.0-\mathrm{g}\) bird is flying horizontally at 2.25 \(\mathrm{m} / \mathrm{s}\) not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 \(\mathrm{cm}\) below the top (Fig. P10.95). The bar is uniform, 0.750 \(\mathrm{m}\) long, has a mass of \(1.50 \mathrm{kg},\) and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away). What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) a uniform solid cylinder; (b) a uniform sphere; ( c) a thin walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\) .

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