Problem 28 - University Physics with Modern Physics

Question

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s}\) . What angular velocity is required to achieve these values?

Step-by-Step Solution

Answer

The required angular velocity is 0.250 rad/s.

Step 1: Define Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is given by the formula \( a_r = \frac{v^2}{r} \), where \(v\) is the tangential velocity and \(r\) is the radius. It can also be expressed as \( a_r = r\omega^2 \), where \( \omega \) is the angular velocity.

Step 2: Derive Equation for Radial Acceleration Without r

To remove \(r\), equate the two expressions for radial acceleration: \(\frac{v^2}{r} = r\omega^2\). Solving for \(r\), we get \(r = \frac{v}{\omega}\). Substitute \(r = \frac{v}{\omega}\) into \( a_r = \omega^2 r\) to obtain: \(a_r = \omega^2 \left(\frac{v}{\omega}\right) = v\omega \). Thus, the equation for radial acceleration without \(r\) is \(a_r = v\omega\).

Step 3: Substitute Known Values

We need to find the angular velocity \(\omega\) given \(a_r = 0.500\,\text{m/s}^2\) and \(v = 2.00\,\text{m/s}\). Substitute these values into the equation \(a_r = v\omega\): \[0.500 = 2.00 \cdot \omega\].

Step 4: Solve for Angular Velocity

Rearrange the equation \(0.500 = 2.00 \cdot \omega\) to find \(\omega\): \[ \omega = \frac{0.500}{2.00} = 0.250\,\text{rad/s} \].

Key Concepts

Centripetal Force

Imagine you are spinning a ball on a string around your head. The force that keeps the ball moving in a circle is called the centripetal force. It acts towards the center of the circle and is responsible for the ball's curved path. Whenever an object moves in a circle or along a curved path, centripetal force ensures that it doesn't fly off in a straight line.
This force is not a separate force of nature but results from other fundamental forces like tension, gravity, or friction. For example:

• A car making a turn experiences centripetal force from the friction between the tires and the road.
• The gravitational pull of the Earth provides the centripetal force needed to keep the Moon in its orbit.
• On a merry-go-round, the structure provides the centripetal force that keeps children safely riding in circular paths.

When we talk about radial acceleration in circular motion, we're referring to how quickly the object's velocity is changing direction, which is directly related to this centripetal force. Without it, circular motion wouldn't be possible!

Angular Velocity

When something rotates or spins, it does so at an angular velocity. Angular velocity measures how quickly an object moves through an angle. Think of it like the speed of a spinning wheel or a rotating fan. It's often measured in radians per second \((rad/s)\).
Angular velocity is crucial when describing rotational motion because it tells us how fast something is spinning. For example:

• When you wash clothes in a washing machine, the drum spins with a certain angular velocity to remove water by centrifugal forces.
• A spinning top maintains its stability because of its angular motion.

The formula relating angular velocity to tangential (or linear) velocity is:\[ v = r \cdot \omega \]where \(v\) is the tangential velocity, \(r\) is the radius of the path, and \(\omega\) is the angular velocity. Changing the angular velocity changes how rapidly different points on the rotating object move through space. A high angular velocity means faster circular movement, while a low angular velocity means slower motion.

Tangential Velocity

Tangential velocity is the speed at which an object moves along its circular path. Imagine a point on the edge of a spinning merry-go-round. This point moves along the circular path, and its speed is the tangential velocity. It's called 'tangential' because it is always directed along the tangent to the path of the circle.
The key thing to remember about tangential velocity is:

• It depends on both the radius of the circular path and the angular velocity.
• Larger radius and higher angular velocity lead to greater tangential velocity.

The relationship between tangential velocity \(v\), radius \(r\), and angular velocity \(\omega\) is:\[ v = r \cdot \omega \]This formula shows how changes in angular velocity or radius affect how quickly the object is moving along the circle. For someone sitting near the edge of a carousel, the ride feels faster because their tangential velocity is much higher than someone closer to the center.
Understanding tangential velocity is crucial for analyzing motions in circular paths and can be seen in various everyday activities, from a spinning CD to the orbits of planets around the sun.