Question

Think \& Calculate A meterstick balances at its center. If an 86-g necklace is suspended from one end of the meterstick, the balance point moves $8.2\mathrm{cm}$ toward that end. (a) Is the mass of the meterstick greater than, less than, or equal to the mass of the necklace? Explain. (b) Find the mass of the meterstick.

Short answer

(a) The mass of the meterstick is greater than the necklace. (b) The mass of the meterstick is 524 g.

Step-by-step solution

Understanding the Problem

The problem requires us to determine whether the mass of a meterstick is greater than, less than, or equal to an 86 g necklace's mass when it is placed at the end of the stick and measures a 8.2 cm shift in the center of balance. Additionally, we need to find the mass of the meterstick.

Application of Torque Equilibrium

In equilibrium, the torque around the pivot (balance point) must be zero. When the necklace is hung 50 cm from the new balance point (100 cm original center - 8.2 cm), it creates a torque. Let the mass of the meterstick be $m$, then we have: $86\times 50=m\times 8.2$.

Solving the Equation

From the torque equilibrium equation $86\times 50=m\times 8.2$, solve for $m$: $$m=\frac{86\times 50}{8.2}\approx 524\text{g}$$

Analysis of Mass Comparison

Since the calculated mass of the meterstick, 524 g, is greater than the necklace's mass of 86 g, the meterstick's mass is greater than the mass of the necklace.

Key concepts

Center of Mass

The center of mass is a crucial concept when dealing with objects and their balance. It represents the point where the mass of an object is perfectly balanced in all directions. In simpler terms, it's like the object's 'balance center'.
If you've ever tried to balance a ruler on your finger, you were finding its center of mass. The object will naturally balance itself when the center of mass is directly above the support point. For the meterstick mentioned in the problem, it balances at its midpoint without any additional mass attached to it.

In practical scenarios, understanding the center of mass helps in predicting how an object will behave when forces are applied. When the necklace is added to one end of the meterstick, the center of mass shifts towards the necklace. This happens because more weight is added to that side, requiring a new balance point closer to the heavier end.

Mass Comparison

Comparing the masses of two objects can help determine how they interact when combined or placed in systems like the meterstick problem. Here, we were tasked to compare the mass of the meterstick to an 86 g necklace.

This can be done using the concept of torque equilibrium, which states that for an object to be balanced, the sum of torques around any pivot must be zero. The torque is calculated as the product of the force (mass times gravity) and the distance from the pivot point. For the necklace and meterstick:

• The torque of the necklace is calculated as the weight of the necklace (86 g) multiplied by its distance (50 cm) from the new center of mass.
• The torque of the meterstick is its mass (m) times the shift in the center (8.2 cm).

Solving this equation, we found the meterstick's mass to be greater, 524 g, indicating a greater influence on the overall balance.

Balance Point

The balance point of an object is where it stays level. It is a critical concept when ensuring an object remains stationary and doesn't tip over.
When the meterstick is balanced at its center, it means both sides have equal mass distribution. Adding a necklace to one side requires the balance point to shift toward it, effectively compensating for the added weight.

The new balance point is determined by finding the exact spot where the torques from each side (necklace and meterstick) cancel each other out. In this problem:

• The initial center was at 50 cm, right at the stick's midpoint.
• Upon adding the necklace, the balance point shifted by 8.2 cm towards the necklace.
• This means the new balance point is now 41.8 cm from the opposite end.

Understanding the process of determining the balance point can be key to solving problems involving multiple forces and weights, ensuring objects maintain equilibrium.