Question

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (\(\textbf{Fig. E8.55}\)). 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?

Short answer

The center of mass will move 0.767 m horizontally; vertically, it doesn't move.

Step-by-step solution

Identify Masses and Distances

We have a horizontal bar with mass \(m_1 = 4.00\, \text{kg}\) and length \(d_1 = 1.50\, \text{m}\). A vertical bar with mass \(m_2 = 3.00\, \text{kg}\) and length \(d_2 = 1.80\, \text{m}\). A ball attached to the end of the vertical bar with mass \(m_3 = 2.00\, \text{kg}\). The point of pivoting is at the hinge joining the two bars.

Initial Horizontal and Vertical Centers of Mass

Calculate the initial center of mass for both horizontal and vertical components. Initially:- The center for the vertical bar is at \((0, d_2/2) = (0, 0.90\, \text{m})\).- The center of mass for the horizontal bar and the ball is at the hinge since they lie along the x-axis at \((d_1/2, 0) = (0.75\, \text{m}, 0)\).

Horizontal and Vertical Center of Mass Shift

When the vertical bar rotates 90 degrees, its center of mass moves from \((0, 0.90)\) to \((0.90, 0)\). Similarly, the ball moves from \((0, 1.80)\) now to \((1.80, 0)\). Calculate new positions for the center of mass of both the bar and the ball.

Calculate Final Centers of Mass

Evaluate the final center of mass using:\[ x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} \]\[ y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} \]Initially (horizontal): \((x_{cm}, y_{cm}) = \left(\frac{4(0.75) + 0 + 0}{9}, \frac{4(0) + 0 + 0}{9}\right) \rightarrow (0.333, 0)\).Finally (horizontal): \((x_{cm}, y_{cm}) = \left(\frac{4(0.75) + 3(0.90) + 2(1.80)}{9}, 0\right) = (1.10, 0)\).

Calculate Horizontal and Vertical Displacement

Calculate the displacement in the horizontal and vertical directions:- \(\Delta x = x_{cm, \text{final}} - x_{cm, \text{initial}} = 1.10 - 0.333 = 0.767\, \text{m}\).- \(\Delta y = y_{cm, \text{final}} - y_{cm, \text{initial}} = 0 - 0 = 0\, \text{m}\).

Key concepts

Horizontal Displacement

The concept of horizontal displacement is crucial when determining how objects move across a plane. In this exercise, we are examining how the center of mass of a machine part shifts as it is rotated from a vertical to a horizontal position.

The initial horizontal position of the center of mass was calculated based on the weighted average positions of the components – the bar and the ball. It initially lies at

• The bar (mass 4.00 kg) has its center of mass at 0.75 m from the pivot.
• The vertical bar (mass 3.00 kg) and the attached ball (2.00 kg) initially have no contribution, as they lie entirely on the vertical axis.

When the system is pivoted to a horizontal alignment, the horizontal center of mass changes significantly:

• The longer bar does not shift, remaining centered at 0.75 m.
• The vertical bar’s center of mass moves to 0.90 m horizontally.
• The ball shifts to 1.80 m horizontally.

After these calculations, the final horizontal center of mass becomes 1.10 m. Thus, the total horizontal displacement is 0.767 m.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on how the mass is distributed in relation to the axis of rotation. In the context of this problem, knowing the moment of inertia would help in understanding the rotational dynamics and the energy involvement as the vertical bar pivots into the horizontal position.

For a uniform bar of length \(L\) and mass \(M\), the moment of inertia \(I\) when it pivots about an end is given by the formula:\[ I = \frac{1}{3}ML^2 \] However, calculating only moment of inertia for the pivoting action in this specific context was not explicitly required for finding the center of mass displacement. Nevertheless, it provides deeper insights when considering further details like:

• How energy distribution varies during the rotation.
• The rotational speed and angular velocity.
• Potential energy changes as the vertical bar pivots downward.

Mechanics Problem Solving

Mechanics problem-solving involves the application of physics principles for analysis and computation. The goal is to break down the problem into manageable parts, apply relevant equations, and interpret results. In this exercise, we used the fundamental notion of center of mass in mechanics.

Here's how we approached the task:

• Identify system components - knowing the mass and distance of each part.
• Compute initial and final positions of centers of mass.
• Apply center of mass formulas: \[x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}\] \[y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}\]
• Analyze results to determine how the overall system behavior, such as horizontal displacement, is affected through the pivot.

Mechanics problem-solving emphasizes precision in calculations and a deep understanding of each component's contribution to the system's dynamics.