91Ó°ÊÓ

Two small disks \(A\) and \(B\), of mass \(2 \mathrm{kg}\) and \(1 \mathrm{kg}\), respectively, may slide on a horizontal and frictionless surface. They are connected by a cord of negligible mass and spin about their mass center \(G . \mathrm{At} t=0\), \(G\) is moving with the velocity \(\overline{\mathrm{v}}_{0}\) and its coordinates are \(\bar{x}_{0}=0, \bar{y}_{0}=1.89 \mathrm{m} .\) Shortly thereafter, the cord breaks and disk \(A\) is observed to move with a velocity \(\mathbf{v}_{A}=(5 \mathrm{m} / \mathrm{s}) \mathrm{j}\) in a straight line and at a distance \(a=2.56 \mathrm{m}\) from the \(y\) axis, while \(B\) moves with a velocity \(\mathrm{v}_{B}=(7.2 \mathrm{m} / \mathrm{s}) \mathrm{i}-(4.6 \mathrm{m} / \mathrm{s}) \mathrm{j}\) along a path intersecting the \(x\) axis at \(\mathrm{a}\) distance \(b=7.48 \mathrm{m}\) from the origin \(0 .\) Determine (a) the initial velocity \(\overline{\mathbf{v}}_{0}\) of the mass center \(G\) of the two disks, \((b)\) the length of the cord initially connecting the two disks, (c) the rate in rad/s at which the disks were spinning about \(G .\)

Step by step solution

01

Understand the Problem

We need to determine three things: (a) the initial velocity of the mass center \(G\), (b) the length of the cord connecting the two disks \(A\) and \(B\), and (c) the spinning rate about \(G\). We'll use principles of conservation of linear momentum and rotational motion for frictionless, non-angular condition.

02

Apply Conservation of Linear Momentum

Before the cord breaks, the linear momentum of the system is the same as after it breaks. We'll use the formula: \[ m_A \mathbf{v}_A + m_B \mathbf{v}_B = (m_A + m_B) \overline{\mathbf{v}}_{0} \]where \(m_A = 2\,\text{kg}\), \(\mathbf{v}_A = (5\,\text{m/s}) \mathbf{j}\), \(m_B = 1\,\text{kg}\), and \(\mathbf{v}_B = (7.2\,\text{m/s}) \mathbf{i} - (4.6\,\text{m/s}) \mathbf{j}\). Solve for the initial velocity \(\overline{\mathbf{v}}_{0}\).

03

Calculate Initial Velocity

Substitute the given values into the momentum conservation equation:\[(2)(0\,\mathbf{i} + 5\,\mathbf{j}) + (1)(7.2\,\mathbf{i} - 4.6\,\mathbf{j}) = (2 + 1)\overline{\mathbf{v}}_0\]Simplify:\[0\,\mathbf{i} + 10\,\mathbf{j} + 7.2\,\mathbf{i} - 4.6\,\mathbf{j} = 3\overline{\mathbf{v}}_0\]\[7.2\,\mathbf{i} + 5.4\,\mathbf{j} = 3\overline{\mathbf{v}}_0\]\[\overline{\mathbf{v}}_0 = 2.4\,\mathbf{i} + 1.8\,\mathbf{j}\,\text{m/s}\].

04

Use Geometry to Find Cord Length

Initially, let the distances of \(A\) and \(B\) from the \(y\)-axis and \(x\)-axis be determined using the final positions. Since \(A\) is \(2.56\,\text{m}\) from the \(y\)-axis and \(B\) is \(7.48\,\text{m}\) from the \(x\)-axis, and given the coordinates are post-separation, use Pythagorean Theorem for the hypotenuse (cord length):\[ \text{Cord Length} = \sqrt{(2.56)^2 + (7.48)^2}\].

05

Calculate Cord Length

Substitute the values into the Pythagorean Theorem:\[\text{Cord Length} = \sqrt{(2.56)^2 + (7.48)^2} = \sqrt{6.5536 + 55.9504} = \sqrt{62.504} \approx 7.906\,\text{m}\].

06

Calculate Spinning Rate

The spinning rate \(\omega\) can be found from the relation \(\omega = \frac{v}{r} \) where \(v\) is the relative velocity between \(A\) and \(B\) and \(r\) is half the cord length (radius of rotation):\[ v = \sqrt{(7.2)^2 + ((-4.6) - 5)^2} = \sqrt{51.84 + 90.25} = \sqrt{142.09} \approx 11.92\,\text{m/s}\omega = \frac{11.92}{7.906/2} \approx 3.02\,\text{rad/s}\].

07

Conclusion

(a) The initial velocity of the mass center \(G\) is \(2.4\,\mathbf{i} + 1.8\,\mathbf{j}\) m/s. (b) The length of the cord is approximately \(7.906\,\text{m}\). (c) The spinning rate is approximately \(3.02\,\text{rad/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Linear Momentum

The conservation of linear momentum is a fundamental concept in physics, particularly crucial in problems involving collisions and separations. It states that the total linear momentum of a closed system remains constant provided no external forces act upon it.

In our exercise, before the cord between the disks breaks, the system's momentum is conserved throughout the event. This means the momentum before the break is equal to the momentum afterward. This principle helps us find the initial velocity of the mass center of the disks, denoted as \(\overline{\mathbf{v}}_0\).

To apply this concept, we use the formula:

\[ m_A \mathbf{v}_A + m_B \mathbf{v}_B = (m_A + m_B) \overline{\mathbf{v}}_0 \]

Here, \(m_A\) and \(m_B\) are the masses, and \(\mathbf{v}_A\), \(\mathbf{v}_B\) are the velocities of disks A and B, respectively. By substituting the given values, we calculate the initial velocity and confirm that linear momentum is conserved. This exercise exemplifies the idea that momentum shifts in a manner reflecting the mass and motion changes in each part of the system.

Rotational Motion

Rotational motion deals with the movement of an object around a central point or axis. This concept is pivotal in understanding how the disks in this problem spin around their common center, \(G\), even as they move linearly.

When analyzing rotational motion, the spinning rate or angular velocity (denoted as \(\omega\)) is an important factor. It describes how fast the object rotates around the center point.

In our scenario, once the cord breaks, each disk continues a rotational effect along their paths. We derive the angular velocity using the formula:

\[ \omega = \frac{v}{r} \]

where \(v\) represents the relative velocity between the two disks and \(r\) is the radius of rotation (half the cord length post-break). By calculating this, we determine the spinning rate at approximate \(3.02\,\text{rad/s}\), indicating how fast they rotate post-separation.

Such calculations help illustrate the elegant interplay between linear and rotational dynamics in systems like these.

Pythagorean Theorem

The Pythagorean Theorem is a crucial tool in physics for calculating distances in problems involving right-angle triangles. It relates the lengths of the sides \(a\), \(b\), and the hypotenuse \(c\) through the formula:

\[ c = \sqrt{a^2 + b^2} \]

In this exercise, once the discs are no longer connected by the cord, we use this theorem to find the initial cord length. Given the distances of disks A and B from the respective axes (2.56 m and 7.48 m post-separation), the theorem helps us calculate the hypotenuse, which represents the length of the cord pre-break.

The calculated cord length is approximately 7.906 m, demonstrating how geometric methods support solving spatial dynamics problems. This theorem's simplicity yet power makes it a staple in analyzing motion scenarios, whether linear or rotational.