Problem 55 - University Physics with Modern Physics

Question

A machine part consists of a thin, uniform \(4.00-\mathrm{kg}\) bar that is 1.50 \(\mathrm{m}\) long, hinged perpendicular to a similar vertical bar of mass 3.00 \(\mathrm{kg}\) and length 1.80 \(\mathrm{m} .\) The longer bar has a small but dense \(2.00-\mathrm{kg}\) ball at one end (Fig. E8.55). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through \(90^{\circ}\) to make the entire part horizontal?

Step-by-Step Solution

Answer

The center of mass moves 0.30 m right and 0.30 m down.

Step 1: Understand the System Before Rotation

The given system consists of two bars and a ball. The first bar ( 4.00 kg, 1.50 m) is horizontal and the second bar ( 3.00 kg, 1.80 m) is vertical with a 2.00 kg ball at the end of the horizontal bar.

Step 2: Find Initial Center of Mass Position

Calculate the center of mass (CM) for each component first. For the horizontal bar (bar1), center of mass is at 0.75 m from the pivot.For the vertical bar (bar2), center of mass is at 0.90 m above the pivot point.For the ball, center of mass is at 1.50 m horizontally.Use: \[ x_{cm,i} = \frac{1}{M} \sum m_i x_i \] \[ y_{cm,i} = \frac{1}{M} \sum m_i y_i \] where \( M = 4 + 3 + 2 = 9 \ \text{kg} \).

Step 3: Calculate Initial CM Coordinates

Calculate the center of mass system coordinates before the rotation: \[ x_{cm,i} = \frac{(4 \times 0.75) + (3 \times 0) + (2 \times 1.50)}{9} = 0.61 \ \text{m} \]\[ y_{cm,i} = \frac{(4 \times 0) + (3 \times 0.90) + (2 \times 0)}{9} = 0.30 \ \text{m} \]

Step 4: Understand the System After Rotation

When the vertical bar rotates counterclockwise 90 degrees, both bars become horizontal.

Step 5: Find Final Center of Mass Position

After rotation, recalculate the mass center.Both bars are now horizontal, so:\[ x_{cm,f} = \frac{(4 \times 0.75) + (3 \times 0.90) + (2 \times 1.50)}{9} = 0.91 \ \text{m} \] \[ y_{cm,f} = 0 \] since both the horizontal bars' CM where vertically coincides with rotational pivot.

Step 6: Calculate the CM Movement Distance

Now compute the displacement.Horizontal displacement: \[ x_{cm,move} = x_{cm,f} - x_{cm,i} = 0.91 - 0.61 = 0.30 \ \text{m} \]Vertical displacement: \[ y_{cm,move} = y_{cm,f} - y_{cm,i} = 0 - 0.30 = -0.30 \ \text{m} \]

Step 7: Conclusion on Center of Mass Movement

The center of mass moves 0.30 m horizontally to the right and 0.30 m vertically downward.

Key Concepts

Torque and Rotational Motion

Rotational motion occurs when an object rotates about an axis. Torque is a measure of how much a force acting on an object causes that object to rotate. You can think of torque as a twist. Imagine opening a door; the force you apply on the door handle is creating torque that causes the door to rotate around its hinges.
To calculate torque, we use the formula:

• \( \tau = r \times F \times \sin(\theta) \)

where \( \tau \) is the torque, \( r \) is the distance to the pivot, or axis of rotation, \( F \) is the force applied, and \( \theta \) is the angle between the force and the arm.In our exercise, the torque is what rotates the previously vertical bar 90 degrees around its hinge to a horizontal position, altering the center of mass of the system. When analyzing systems with rotation, it is crucial to understand the effects of individual masses and their distances from the pivot on the overall torque and rotational motion.

Physical Pendulum

A physical pendulum refers to a rigid body swinging about a pivot point. Unlike a simple pendulum, which is usually idealized as a point mass at the end of a massless string, a physical pendulum includes the distribution of mass in its calculations. The distribution influences its oscillation period and stability.For instance, if you look at the longer bar with the ball, rotating to become horizontal, it acts as a physical pendulum when it swings about the pivot. The location of the pivot and how mass is distributed around it are key considerations in such systems.
To determine how a physical pendulum behaves, we use its moment of inertia, \( I \), which depends on how its mass is spread out. The angular frequency of small oscillations is given by:

• \( \omega = \sqrt{\frac{mgd}{I}} \)

where \( m \) is the mass, \( g \) is the acceleration due to gravity, \( d \) is the distance from the pivot to the center of mass, and \( I \) is the moment of inertia. Understanding the dynamics of physical pendulums is crucial in many mechanical applications, especially in engineering and physics.

Equilibrium in Physics

Equilibrium in physics refers to a state where the sum of all forces and the sum of all torques acting on an object are zero. This means the object either remains at rest or moves with constant velocity, and there is no net change in rotational motion.
For the machine part in the exercise, equilibrium concepts underpin the changes observed when the bar is rotated. Initially, the system is balanced vertically. When altered, the equilibrium state shifts because of the motion enacted by the torque on the pivot. An object is in static equilibrium when it is not moving. In dynamic equilibrium, even though the object moves, there is no change in its motion. In our problem, the equilibrium changes when external torque rotates the bar, thus altering the center of mass, but eventually, it stabilizes horizontally. This reflects how equilibrium is maintained under different orientations in mechanical systems.