Problem 27 - College Physics

Question

A little boy jumps onto a small merry-go-round (radius of \(2.00 \mathrm{~m}\) ) in a park and rotates for 2.30 s through an arc length distance of \(2.55 \mathrm{~m}\) before coming to rest. If he landed (and stayed) at a distance of \(1.75 \mathrm{~m}\) from the central axis of rotation of the merry-go-round, what was his average angular speed and average tangential speed?

Step-by-Step Solution

Answer

Average angular speed: \( \omega = 0.64 \, \text{rad/s} \); Average tangential speed: \( v = 1.12 \, \text{m/s} \).

Step 1: Identify Angular Displacement

First, we calculate the angular displacement \( \theta \) using the formula \( \theta = \frac{s}{r} \), where \( s = 2.55 \, \text{m} \) is the arc length and \( r = 1.75 \, \text{m} \) is the radius at which the boy is sitting. So, \( \theta = \frac{2.55}{1.75} \, \text{rad} \).

Step 2: Calculate Angular Speed

With the angular displacement \( \theta \) calculated, the average angular speed \( \omega \) is given by \( \omega = \frac{\theta}{t} \), where \( t = 2.30 \, \text{s} \) is the time. Substitute the values \( \theta = \frac{2.55}{1.75} \, \text{rad} \) and \( t = 2.30 \, \text{s} \) to find \( \omega \).

Step 3: Determine Tangential Speed

Once the average angular speed \( \omega \) is known, the average tangential speed \( v \) can be found using the relationship \( v = r \cdot \omega \). Here, \( r = 1.75 \, \text{m} \) and \( \omega \) is the value calculated in the previous step.

Key Concepts

Understanding Angular Displacement

Angular displacement is a key concept in rotational motion. It measures the angle through which an object moves on a circular path. Think of it as the rotational equivalent of linear distance. Imagine you are drawing an arc on a circle, the angle of this arc at the center of the circle is what we call angular displacement.

Mathematically, it's calculated using the formula:

• \( \theta = \frac{s}{r} \)

where \(s\) is the arc length, and \(r\) is the radius of the circle. In the example of the boy on the merry-go-round, his angular displacement can be calculated because we know the distance he covered along the arc, which is 2.55 m, and his distance from the center of the merry-go-round, which is 1.75 m.

• This tells us that the boy has an angular displacement of \(\frac{2.55}{1.75}\) radians.

So, when you look at the entire rotation, focus on the angle measure rather than the surface distance. It's a different but equally important way to quantify movement.

Diving Into Angular Speed

Angular speed is like the speedometer of a car, but for rotations. It defines how fast an object rotates or spins around a central point. In other words, it tells us the rate of change of angular displacement with respect to time.

To find angular speed, we use the formula:

• \( \omega = \frac{\theta}{t} \)

where \(\theta\) is the angular displacement and \(t\) is the time taken. So in our example, after finding the angular displacement of the boy on the merry-go-round which is \(\frac{2.55}{1.75}\) radians, we divide it by the time he spent rotating, which is 2.30 seconds. This gives us his average angular speed.

• Angular speed tells you how quickly something is spinning, and it's measured in radians per second.

Remember, while angular speed tells how fast the angle is changing, it doesn't give details about how fast any particular point is moving through space.

Exploring Tangential Speed

Tangential speed refers to the linear speed of something moving along the edge of a circular path. Imagine standing on a merry-go-round. As it spins, you're moving around its edge, and the speed at which you're moving is the tangential speed. It's like measuring how fast you can run along the perimeter of the circle the merry-go-round traces out.

To calculate this, we use the relationship:

• \( v = r \cdot \omega \)

where \(r\) is the radius – in this scenario, the boy's distance from the center of rotation, 1.75 m, and \(\omega\) is the average angular speed we calculated before.

• This formula integrates rotational motion with linear motion components, showing how they connect.

Understanding this is crucial because it directly relates rotational speed to actual speed at the edge of a circle, which is practically experienced in many everyday situations, like car wheels on the road or children playing on roundabouts.