Question

A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Short answer

The tangential velocity is approximately 8.21 m/s.

Step-by-step solution

Understanding the Relationship between Kinetic Energy and Moment of Inertia

The kinetic energy \( K \) of a rotating object is given by the formula: \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. For a solid sphere rotating about its diameter, \( I = \frac{2}{5} m r^2 \), where \( m \) is mass and \( r \) is the radius.

Solving for Angular Velocity \( \omega \)

Given \( K = 236 \text{ J} \), \( m = 28.0 \text{ kg} \), and \( r = 0.380 \text{ m} \), we substitute the moment of inertia into the kinetic energy formula: \( 236 = \frac{1}{2} \left(\frac{2}{5} \cdot 28.0 \cdot 0.380^2\right) \omega^2 \). First, calculate the moment of inertia: \( I = \frac{2}{5} \cdot 28.0 \cdot 0.380^2 = 1.01232 \text{ kg} \cdot \text{m}^2 \). Substitute back to find \( \omega \): \( 236 = \frac{1}{2} \times 1.01232 \times \omega^2 \). Solving for \( \omega \), we find \( \omega \approx 21.61 \text{ rad/s} \).

Calculating Tangential Velocity

Once \( \omega \) is known, the tangential velocity \( v_t \) of a point at the edge of the sphere is related by the formula \( v_t = \omega r \). Substituting in the known values: \( v_t = 21.61 \cdot 0.380 = 8.2118 \text{ m/s} \).

Key concepts

Kinetic Energy

Kinetic energy in rotational dynamics refers to the energy an object possesses due to its rotation. Unlike linear kinetic energy, which depends on an object’s mass and velocity, rotational kinetic energy depends on the object's moment of inertia and angular velocity.
The formula used is \[ K = \frac{1}{2} I \omega^2 \] where:

• \( K \) is the kinetic energy.
• \( I \) is the moment of inertia, which captures how mass is distributed relative to the axis of rotation.
• \( \omega \) represents the angular velocity, or how fast the object is rotating.

Understanding this relationship helps calculate how much energy is stored in a rotating system—key for deciphering other dynamic properties like the sphere in our example.

Moment of Inertia

The moment of inertia is a measure that reflects how mass is positioned relative to the axis of rotation. Think of it as the rotational equivalent of mass in linear motion. It's crucial because it influences how easily an object can be put into rotational motion or stopped.
For different shapes and axes, the moment of inertia will vary. Focusing on a solid sphere rotating about its diameter, the moment of inertia is given by \[ I = \frac{2}{5} m r^2 \] where:

• \( m \) is the mass of the sphere.
• \( r \) is the radius of the sphere.

This formula helps describe how the mass distribution affects the sphere’s rotational characteristics, which is evident in our example where we calculated \( I \approx 1.01232 \text{ kg} \cdot \text{m}^2 \).

Angular Velocity

Angular velocity represents how quickly an object rotates around an axis. It is expressed in radians per second (\( \text{rad/s} \)), making it essential for understanding rotational motion just as regular velocity describes straight-line motion.

• It gives insight into the speed of rotation.
• Helps in determining other dynamic variables like tangential velocity.

From the kinetic energy equation \( K = \frac{1}{2} I \omega^2 \), solving for \( \omega \) helps us determine the rate at which the sphere spins. For our sphere example, we found \( \omega \approx 21.61 \text{ rad/s} \), indicating a fairly rapid spin around its axis.

Tangential Velocity

Tangential velocity refers to the linear speed of a point on the outer surface of a rotating object. It's what you would perceive as speed if you were standing on that edge point as the sphere spins.
Calculating tangential velocity involves multiplying angular velocity by the radius of the rotation: \[ v_t = \omega r \].This formula shows the direct relationship between how fast the sphere spins and how large its radius is.

• \( v_t \) provides a more intuitive understanding of motion, as it's a linear speed.
• Helps connect rotational dynamics with linear motion concepts.

In our scenario, we calculated \( v_t \approx 8.2118 \, \text{m/s} \), illustrating how quickly a point on the sphere's edge moves.