Question

A lightweight plastic rod has a mass of \(1.0 \mathrm{~kg}\) attached to one end and a mass of \(1.5 \mathrm{~kg}\) attached to the other end. The rod has a length of \(0.80 \mathrm{~m}\). How far from the \(1.0-\mathrm{kg}\) mass should a string be attached to balance the rod?

Short answer

The string should be attached 0.48 m from the 1.0 kg mass.

Step-by-step solution

Identify the Problem

We need to find the point along the rod where it can be balanced. This involves finding the center of mass for two point masses on a rod. Our objective is to determine how far from the 1.0 kg mass the center of mass is located.

Use the Center of Mass Formula

For two masses, the center of mass, \( x \), along a rod can be calculated using the formula: \( x = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \) where \( m_1 = 1.0 \text{ kg} \) and \( m_2 = 1.5 \text{ kg} \). We can assume \( x_1 = 0 \), the position of the 1.0 kg mass, and \( x_2 = 0.80 \text{ m} \), the position of the 1.5 kg mass.

Plug Values into the Formula

Substitute the values into the center of mass formula: \[x = \frac{(1.0 \text{ kg} \times 0 \text{ m}) + (1.5 \text{ kg} \times 0.80 \text{ m})}{1.0 \text{ kg} + 1.5 \text{ kg}} = \frac{0 + 1.2}{2.5} = 0.48 \text{ m}\]

Interpret the Result

The center of mass is \(0.48\text{ m}\) from the 1.0 kg mass. Thus, the string should be attached \(0.48\text{ m}\) from the 1.0 kg mass to balance the rod.

Key concepts

Balanced Rod

A balanced rod is an object that remains level and steady because its center of mass is perfectly supported. Imagine holding a rod horizontally with weights at each end. If you attach a string exactly at the center of mass, the rod will not tip. This happens because the torques on either side of the string are equal, keeping the rod in equilibrium.

• The rod stays level if the net torque around the point of support is zero.
• In our exercise, the rod needs support where the string balances its weight distribution.

By understanding this equilibrium, we can predict where to add supports or how to position weights for balanced structures.

Point Masses

Point masses are simplified representations of objects where all the mass is concentrated at a single point. They help us calculate the center of mass easily without dealing with the complexities of real shapes.

• In calculations, point masses simplify the math by ignoring the dimensions of objects.
• We placed the rod's two weights as point masses at the ends to apply the center of mass formula.

This simplification makes it straightforward to determine balance points, especially in physics problems involving rods or beams.

Physics Problem-Solving

Physics problem-solving involves breaking down complex situations into manageable steps. To solve such problems efficiently, follow these strategies:

• Identify what is being asked: Determine the exact question — a crucial step for correct solutions.
• Use relevant formulas: Here, the center of mass formula is essential to find the balancing point.
• Substitute known values: place the quantities you know into the formula—easier with numbers from the given problem.
• Interpret results: Understand what the mathematical outcome means in practical terms.

These steps make complex physics topics approachable and ensure calculation accuracy.

Mass Distribution

Mass distribution refers to how mass is spread out in an object, affecting balance and motion. More mass on one side of an object means that side will be heavier, thus more challenging to balance without knowing its mass distribution.

• The more equally distributed the mass, the easier to find a balance.
• In the rod example, the masses are not equal, so the center of mass isn't at the midpoint.

Understanding mass distribution is crucial when designing and managing balanced systems such as beams, rods, or even seesaws.