Problem 54 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Two blocks, with masses \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}},\) are connected to each other and to a central post by cords as shown in Fig. \(5-46 .\) They rotate about the post at frequency \(f\) (revolutions per second) on a frictionless horizontal surface at distances \(r_{\mathrm{A}}\) and \(r_{\mathrm{B}}\) from the post. Derive an algebraic expression for the tension in each segment of the cord (assumed massless).

Step-by-Step Solution

Answer

Express tensions as: \(T_{\text{AB}} = m_{\text{A}} r_{\text{A}} (2\pi f)^2\) and \(T_{\text{B}} = m_{\text{A}} r_{\text{A}} (2\pi f)^2 + m_{\text{B}} r_{\text{B}} (2\pi f)^2\).

Step 1: Identify Forces Acting on the Blocks

Each block experiences a centripetal force due to its rotation around the post. This force is provided by the tension in the cord connecting the block to the post and the adjacent block.

Step 2: Write the Expression for Centripetal Force

For block A, the centripetal force is provided by the tension in the cord connecting it to block B. The centripetal force is given by the equation \(F_{cA} = m_{ ext{A}} r_{ ext{A}} (2\pi f)^2\). For block B, the centripetal force is provided by the tension in the cord connecting it to the post. The centripetal force is given by the equation \(F_{cB} = m_{ ext{B}} r_{ ext{B}} (2\pi f)^2\).

Step 3: Determine Tension in the Cord Between Block A and Block B

The tension in the cord between block A and block B, which we'll call \(T_{ ext{AB}}\), is equal to the centripetal force acting on block A due to the rotation, provided by the cord tension: \[T_{ ext{AB}} = m_{ ext{A}} r_{ ext{A}} (2\pi f)^2.\]

Step 4: Determine Tension in the Cord Connected to Block B and the Post

The tension in the cord connected between block B and the post, \(T_{ ext{B}}\), must account for both the centripetal force required for block B and the force due to the tension in the section of the cord between block A and block B. Thus:\[T_{ ext{B}} = T_{ ext{AB}} + m_{ ext{B}} r_{ ext{B}} (2\pi f)^2.\] By substituting the expression for \(T_{ ext{AB}}\), we get:\[T_{ ext{B}} = m_{ ext{A}} r_{ ext{A}} (2\pi f)^2 + m_{ ext{B}} r_{ ext{B}} (2\pi f)^2.\]

Key Concepts

Tension in Cord

When you think about tension in a cord, it's like considering the pulling force that's stretching the cord. Imagine a rope tied to a weight that's being spun around in circles. The tension is what keeps the weight moving in that circle without falling away.

In our exercise scenario, two blocks are attached to cords which in turn are connected to a central post. Thanks to the rotation, tension builds up in these cords.

To find the tension in the cord between block A and block B, we acknowledge that the force keeping block A moving in a circular path—its centripetal force—is exactly what the tension in this cord must match. That's why we see the equation:
\[T_{AB} = m_A r_A (2\pi f)^2\]
showing that tension depends on both the mass of block A and its distance from the central post.

Next, the tension in the cord that’s attaching block B to the central post involves both the centripetal force needed for block B and the tension already calculated for block A

\[T_{B} = m_A r_A (2\pi f)^2 + m_B r_B (2\pi f)^2\]

• Note how each block’s mass and radius factor into the amount of tension.
• The faster the blocks spin (higher frequency), the greater the tension.

Frictionless Surface

The term "frictionless surface" might sound like it's something out of science fiction, but it's a crucial concept for simplifying problems involving motion.
In our context, a frictionless surface means that there's no resisting force acting against the blocks' movement.

This lack of friction ensures that:

• The only force needed to maintain the motion is the centripetal force generated by the cords.
• We don't have to calculate or include any force of friction in our tension equations.

Without friction, the blocks can smoothly rotate around the post as their paths are not hindered by any opposing forces.

This makes it easier to focus purely on how the rotation and tension interact, providing clear insight into circular motion dynamics.
Remember, frictionless surfaces are theoretical but very effective for learning purposes. They eliminate complexities, allowing us to clearly see the underlying principles at work.

Rotational Motion

Rotational motion involves objects moving in a circle around a central point. Think of a carousel in a park. As the ride spins, the horses move in a circular path around the center.

In the exercise, blocks rotate around a central post, depicting rotational motion beautifully. Key elements to consider:

• Centripetal force: Required to keep an object moving in a circle, always directed toward the center. For the blocks, this is provided by the tension in the cord.
• Frequency of rotation: Denoted as \(f\), it represents how many turns the blocks make around the post in one second. More spins mean higher frequency and greater required centripetal force.
• Radius: Distance from the central point (post) to the rotating blocks, it influences the amount of tension needed in the cord.

In rotational motion, all these factors interconnect. By understanding how they work together, students can comprehend the dynamics of circular paths easily and efficiently.