Problem 4 - College Physics

Question

A bowling ball with a radius of \(15.0 \mathrm{~cm}\) travels down the lane so that its center of mass is moving at \(3.60 \mathrm{~m} / \mathrm{s} .\) The bowler estimates that it makes about 7.50 complete revolutions in 2.00 seconds. Is it rolling without slipping? Prove your answer, assuming that the bowler's quick observation limits answers to two significant figures.

Step-by-Step Solution

Answer

Yes, the ball is rolling without slipping as the calculated and given speeds are approximately equal.

Step 1: Calculate the Angular Speed of the Bowling Ball

First, we need to determine the angular speed \( \omega \) of the bowling ball. We know that the ball makes 7.50 revolutions in 2.00 seconds. Since 1 revolution is \( 2\pi \) radians, the total angle \( \theta \) covered is \( 7.50 \times 2\pi \) radians. The formula for angular speed is \( \omega = \frac{\theta}{t} \), where \( t \) is time. Thus, \( \omega = \frac{15\pi}{2} \approx 23.56 \, \text{rad/s}\).

Step 2: Find Linear Speed Due to Rotation

The linear speed \( v \) due to rotation can be found using the formula \( v = r\omega \), where \( r \) is the radius of the ball, \( 0.15 \, \text{m} \). Substituting the values, we get \( v = 0.15 \times 23.56 \approx 3.53 \, \text{m/s}\).

Step 3: Compare Actual Speed with Rolling Speed

We compare the calculated linear speed \( 3.53 \, \text{m/s} \) with the center of mass speed \( 3.60 \, \text{m/s} \) given in the problem. For the ball to roll without slipping, these speeds should be approximately equal.

Step 4: Determine if the Ball is Rolling Without Slipping

Since the calculated linear speed due to rotation \( 3.53 \, \text{m/s} \) is very close to the actual center of mass speed \( 3.60 \, \text{m/s} \), and considering the precision allowed (two significant figures), we conclude that the ball is rolling without slipping.

Key Concepts

Angular Speed

Angular speed describes how quickly an object rotates or spins around its axis. In our bowling ball problem, angular speed is crucial to understanding the ball's rotation as it travels down the lane. Angular speed is usually denoted by the Greek letter \(\omega\) and is measured in radians per second (rad/s).
- **To calculate angular speed**, you first need to know how many rotations or revolutions an object makes. In this problem, the bowling ball completes 7.50 revolutions in 2.00 seconds.
- Each revolution of the ball corresponds to a complete circle, or \(2\pi\) radians.
- Thus, the total angle the ball covers is \(7.50 \times 2\pi\) radians.Using the formula for angular speed, \(\omega = \frac{\theta}{t}\), where \(\theta\) is the angle in radians and \(t\) is the time in seconds, we find \(\omega = \frac{15\pi}{2} \approx 23.56\, \text{rad/s}\). This calculation helps in understanding how fast the ball spins as it rolls.

Linear Speed

Linear speed is the rate at which an object moves along a path. It's crucial when analyzing rolling objects like our bowling ball.
- In essence, linear speed tells us how fast the center of mass of the ball travels down the lane.
- The linear speed due to rotation can be linked to the angular speed using the equation \(v = r\omega\), where \(r\) is the radius of the ball and \(\omega\) is the angular speed calculated earlier.
Knowing the radius of the ball is 0.15 meters, we can calculate:
\[v = 0.15 \times 23.56 \approx 3.53\, \text{m/s}\]This linear speed due to rotation needs to closely match the actual linear speed of the ball's center of mass, calculated as 3.60 m/s, to check if it rolls without slipping.

Rolling Without Slipping

When an object rolls without slipping, there is no relative motion between the point of contact and the surface. In simpler terms, the ball spins just right to match its movement down the lane.
- For rolling without slipping, the linear speed of the outer surface due to rotation must be equal to the speed of the ball's center of mass.
- In our bowling ball example, the calculated linear speed due to rotation was 3.53 m/s, closely matching the observed center of mass speed of 3.60 m/s.
Given the small difference and the consideration of two significant figures (which accounts for minor discrepancies), we conclude that the bowling ball is indeed rolling without slipping. This concept ensures that the ball rotates correctly relative to its motion, providing a smooth path with no skidding along the lane.