Question

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

The tumbler has 138.6 J total kinetic energy, with 33.33% being rotational.

Step-by-step solution

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:\[ I = \frac{1}{2} \times 75 \times (0.5)^2 = 9.375 \text{ kg m}^2 \].

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): \[ \omega = 0.50 \times 2\pi = \pi \text{ rad/s} \].

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 \( \omega \):\[ KE_{rot} = \frac{1}{2} \times 9.375 \times (\pi)^2 = 46.2 \text{ J} \].

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 \( KE_{trans} \):\[ KE_{trans} = \frac{1}{2} \times 75 \times (0.5\pi)^2 = 92.4 \text{ J} \].

Calculate Total Kinetic Energy

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

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:\[ \frac{46.2}{138.6} \times 100\% \approx 33.33\% \].

Key concepts

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.