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We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 kg and diameter 1.0 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 138.6 J total kinetic energy, with 33.33% being rotational.

Step by step solution

01

Calculate the Moment of Inertia

For a uniform solid cylinder, the moment of inertia \( I \) about its central axis is given by the formula \( I = \frac{1}{2} m r^2 \). Here, the mass \( m \) is 75 kg, and since the diameter is 1.0 m, the radius \( r \) is 0.5 m. Plugging in these values, we get:\[ I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \text{ kg m}^2 \].
02

Determine the Angular Velocity

The angular velocity \( \omega \) is related to the frequency of rotation. Given that the tumbler rolls at 0.50 revolutions per second, we convert this to radians per second (since 1 revolution is \(2\pi\) radians): \[ \omega = 0.50 \times 2\pi = \pi \text{ rad/s} \].
03

Calculate Rotational Kinetic Energy

The rotational kinetic energy \( KE_{rot} \) is given by the formula \( KE_{rot} = \frac{1}{2} I \omega^2 \). Using the previously calculated \( I \) and \( \omega \):\[ KE_{rot} = \frac{1}{2} \times 9.375 \times (\pi)^2 = 46.2 \text{ J} \].
04

Calculate Translational Kinetic Energy

The translational kinetic energy \( KE_{trans} \) is given by the formula \( KE_{trans} = \frac{1}{2} m v^2 \). The linear velocity \( v \) can be found using the relation \( v = r \omega \), so \( v = 0.5 \times \pi \). Now, calculating \( KE_{trans} \):\[ KE_{trans} = \frac{1}{2} \times 75 \times (0.5\pi)^2 = 92.4 \text{ J} \].
05

Calculate Total Kinetic Energy

The total kinetic energy \( KE_{total} \) is the sum of the rotational and translational kinetic energies:\[ KE_{total} = KE_{rot} + KE_{trans} = 46.2 + 92.4 = 138.6 \text{ J} \].
06

Calculate the Percent of Rotational Kinetic Energy

To find the percentage of the total kinetic energy that is rotational, use the formula:\[ \frac{KE_{rot}}{KE_{total}} \times 100\% \]Calculating this gives:\[ \frac{46.2}{138.6} \times 100\% \approx 33.33\% \].

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Key Concepts

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

Moment of Inertia
The moment of inertia is a measure of how much an object resists changes in its rotation.Think of it as the rotational equivalent of mass for linear motion.
For a uniform solid cylinder, such as our gymnastic tumbler, the moment of inertia is calculated using the formula:
  • \( I = \frac{1}{2} m r^2 \)
Here:
  • \( m \) is the mass, which in this case, is 75 kg.
  • \( r \) represents the radius. Given a diameter of 1.0 m, the radius \( r \) becomes 0.5 m.
Substituting these values into the formula, we calculate:
  • \( I = \frac{1}{2} \times 75 \times 0.5^2 = 9.375 \) kg\( \cdot \)m\(^2\)
This value of moment of inertia tells us how much rotational force is needed for the tumble to spin about its axis.
Angular Velocity
Angular velocity describes how fast an object is rotating. In simple terms, it shows how much angle is covered per unit time.
In the exercise for the gymnastics tumbler, he rolls at a speed of 0.50 revolutions per second. But to use this value in calculations, we need to express it in radians per second because 1 revolution equals \( 2\pi \) radians.
The conversion is calculated as follows:
  • \( \omega = 0.50 \times 2\pi = \pi \; \, \text{rad/s} \)
By converting the speed to radians per second, we can easily integrate this value into other calculations regarding rotational dynamics.
Translational Kinetic Energy
Translational kinetic energy refers to the energy of an object due to its linear motion.
For the gymnastic tumbler, you need to understand how the forward motion contributes to the overall kinetic energy. Using the relation between linear velocity and angular speed, we can explore this concept.
Firstly, to find the linear velocity \( v \) of the cylinder, use the relationship \( v = r \omega \). Substituting the known values gives:
  • \( v = 0.5 \times \pi \)
The translational kinetic energy can then be computed by the formula:
  • \( KE_{trans} = \frac{1}{2} m v^2 \)
Plugging in the linear velocity, we have:
  • \( KE_{trans} = \frac{1}{2} \times 75 \times (0.5\pi)^2 = 92.4 \text{ J} \)
This express the energy portion of the tumbler due to moving from one point to another in space.

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Most popular questions from this chapter

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg \(\cdot\) m\(^2\). You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is 70.0 kg. Since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip. What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?

A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_x\) and \(a_z\) are approximately zero and \(v_x\) and \(\omega_z\) are approximately constant. Rolling without slipping means \(v_x = r\omega_z\) and \(a_x = r\alpha_z\) . If an object is set in motion on a surface \(without\) these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_0\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_k\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_x\) of the center of mass and \(a_z\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_z = \omega_0\) but \(v_x =\) 0. Rolling without slipping sets in when \(v_x = r\omega_z\) . Calculate the \(distance\) the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

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