A 30 g ball rolls around a 40 cm -diameter L-shaped track, shown in the figure, (Figure 1) at 50 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected. Hint: The track exerts more than one force on the ball.

Let's analyze the problem systematically to calculate the net force exerted on the ball by the track. We'll break it down into several steps, keeping in mind that the ball is undergoing circular motion, and the forces involved are due to both radial and vertical components.

### Step 1: Identifying the relevant forces

1. **Centripetal Force**: The ball is rolling in a circular path, which requires a centripetal force to keep it in circular motion. This force will be provided by the track in the horizontal direction.
2. **Gravitational Force**: The weight of the ball acts vertically downward, \( F_g = mg \), where \( g \) is the acceleration due to gravity.

### Step 2: Given information

- Mass of the ball, \( m = 30 \, \text{g} = 0.03 \, \text{kg} \)
- Diameter of the track, \( d = 40 \, \text{cm} = 0.40 \, \text{m} \), so the radius of the track is \( r = \frac{d}{2} = 0.20 \, \text{m} \)
- Rotational speed of the ball, \( f = 50 \, \text{rpm} \) (revolutions per minute)

### Step 3: Convert angular velocity to rad/s

To calculate the centripetal force, we first need the angular velocity in radians per second. We can convert from rpm (revolutions per minute) to radians per second using the formula:

\[
\omega = 2\pi \times \frac{f}{60}
\]

Substituting \( f = 50 \, \text{rpm} \):

\[
\omega = 2\pi \times \frac{50}{60} = 5.24 \, \text{rad/s}
\]

### Step 4: Calculate the centripetal force

The centripetal force is given by:

\[
F_{\text{centripetal}} = m \omega^2 r
\]

Substituting the values:

\[
F_{\text{centripetal}} = 0.03 \times (5.24)^2 \times 0.20 = 0.03 \times 27.45 \times 0.20 = 0.165 \, \text{N}
\]

### Step 5: Gravitational force

The gravitational force acting on the ball is:

\[
F_g = mg = 0.03 \times 9.8 = 0.294 \, \text{N}
\]

### Step 6: Net force

The track exerts a force on the ball to provide both the centripetal force (in the horizontal direction) and a normal force to counteract the gravitational force (in the vertical direction). These two forces are perpendicular to each other, so the magnitude of the net force is given by the Pythagorean theorem:

\[
F_{\text{net}} = \sqrt{F_{\text{centripetal}}^2 + F_g^2}
\]

Substituting the values:

\[
F_{\text{net}} = \sqrt{(0.165)^2 + (0.294)^2} = \sqrt{0.0272 + 0.0864} = \sqrt{0.1136} = 0.337 \, \text{N}
\]

### Final Answer

The magnitude of the net force that the track exerts on the ball is approximately **0.337 N**.

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Do you want more details or have any questions?

Here are 5 related questions that expand on the concepts:

1. How would the net force change if the ball’s speed were doubled?
2. What would happen to the net force if the mass of the ball increased?
3. How does the centripetal force depend on the radius of the circular track?
4. Can you explain the difference between centripetal force and centrifugal force in this context?
5. What role does rolling friction play if it was not neglected?

**Tip**: Always convert rotational units like rpm into rad/s to calculate forces related to circular motion!

A 30 g ball rolls around a 40 cm-diameter L-shaped track at 50 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected.

Learn how to calculate the net force on a 30 g ball rolling on a 40 cm-diameter L-shaped track at 50 rpm. This problem involves concepts of circular motion, centripetal force, and gravitational force. Suitable for students in Grades 11-12.

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