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A cube of mass \(m\) slides without friction at speed \(v_{0} .\) It undergoes a perfectly elastic collision with the bottom tip of a rod of length \(d\) and mass \(M=2 m .\) The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity - both speed and direction-after the collision?

Short Answer

Expert verified
Velocity of the cube after collision is \(v_1\), facing upwards.

Step by step solution

01

Conservation of Momentum

Before the collision, the box moves with speed \(v_0\) while the rod is at rest. After the collision, the box and rod both move. Let \(v_1\) be the final speed of the box and let \(ω\) be the final angular speed of the rod. We'll take upward and clockwise directions as positive. From the conservation of momentum: \(mv_0 = mv_1 + M*dω/2\).
02

Conservation of Kinetic Energy

In the collision, the kinetic energy is also conserved. The initial kinetic energy was all in the box, and the final kinetic energy is shared by the box and the rod. The kinetic energy of the box is \(1/2 m v_1^2\), and the kinetic energy of the rotating rod is \(1/2 I ω^2\), where \(I\) is the moment of inertia of the rod about its center of mass, given by \(M*d^2/12\). So: \(1/2 m v_0^2 = 1/2 m v_1^2 + 1/2 *(M*d^2/12) * ω^2.\)
03

Solve the Equations

We now have two equations from steps 1 and 2, one from the conservation of momentum and one from the conservation of kinetic energy. We can solve these equations simultaneously to find \(v_1\) and \(ω\). After finding \(v_1\), we can conclude that the post-collision speed of the cube is equal to \(v_1\), facing upwards.

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Key Concepts

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

Conservation of Momentum
One fundamental principle in physics is the conservation of momentum, which states that the total momentum of an isolated system remains constant if no external forces are acting on it. Momentum, a property of all moving objects, is defined as the product of an object's mass and its velocity. In our exercise, before the cube collides with the rod, it has a linear momentum of \(m v_0\).

The collision is an internal interaction, so the system's total momentum must be conserved. After the collision, the cube's momentum changes, and the rod begins to rotate, gaining angular momentum. The final linear momentum of the cube and the final angular momentum of the rod must add up to the initial momentum of the cube. The angular momentum of the rod is related to its linear momentum by the lever arm (half its length, since the pivot is at the center), forming the equation \(mv_0 = mv_1 + M \frac{d\omega}{2}\).

Understanding this concept is crucial as it sets the foundation for solving elastic collision problems and is widely applied in various fields of science and engineering.
Conservation of Kinetic Energy
In an elastic collision, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is the energy an object possesses due to its motion, calculated as \(\frac{1}{2}mv^2\) for linear motion and \(\frac{1}{2}I\omega^2\) for rotational motion, where \(I\) is the moment of inertia and \(\omega\) is the angular speed.

In our scenario, the kinetic energy of the cube before the collision is \(\frac{1}{2} m v_0^2\). After the collision, the kinetic energy is distributed between the moving cube and the rotating rod. Writing this conservation equation allows us to express the post-collision scenario as \(\frac{1}{2} m v_0^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (M\frac{d^2}{12}) \omega^2\). By solving the conservation of momentum and kinetic energy equations together, we determine the system's final state post-collision. This concept is key to analyzing and predicting the outcomes of various physical systems where collisions are involved.
Angular Speed
When we talk about angular speed, we're discussing how fast an object is rotating. Specifically, it's the rate at which the object covers an angle, typically measured in radians per second. In the context of our exercise, the angular speed \(\omega\) represents how quickly the rod is rotating about its pivot point after the collision.

The faster the angular speed, the greater the influence it has on the system's kinetic energy due to rotation. Angular speed ties in closely with both the conservation of momentum and the conservation of kinetic energy, as seen in the equations provided in the steps. It's the rotational equivalent of linear speed but applied to rotational motion which is crucial for understanding systems that include spinning or revolving components.
Moment of Inertia
The moment of inertia (\(I\)) is essentially a measure of an object's resistance to changes in its rotation about an axis. It depends not only on the object's mass but also on how that mass is distributed relative to the axis of rotation. For our rod, pivoted at its center, the moment of inertia can be calculated using the formula \(I = M\frac{d^2}{12}\), where \(M\) is the mass of the rod, and \(d\) is its length.

The moment of inertia is akin to mass in linear motion; it plays a key role in the equations used to calculate the conservation of kinetic energy during rotational motion. The distribution of mass affects how an object will rotate when subjected to a force, influencing its angular acceleration – a critical component to master when dealing with any rotational mechanics problems.

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Most popular questions from this chapter

Vector \(\vec{A}=3 \hat{\imath}+\hat{\jmath}\) and vector \(\vec{B}=3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k}\)a. What is the cross product \(\vec{A} \times \vec{B} ?\) b. Show vectors \(\vec{A}, \vec{B},\) and \(\vec{A} \times \vec{B}\) on a three- dimensional coordinate system.

A long, thin rod of mass \(M\) and length \(L\) is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are (a) the angular velocity and (b) the speed of the tip of the rod?

A solid sphere of radius \(R\) is placed at a height of \(30 \mathrm{cm}\) on a \(15^{\circ}\) slope. It is released and rolls, without slipping, to the bottom. a. From what height should a circular hoop of radius \(R\) be released on the same slope in order to cqual the sphere's speed at the bottom? b. Can a circular hoop of different diameter be released from a height of \(30 \mathrm{cm}\) and match the sphere's speed at the bottom? If so, what is the diameter? If not, why not?

Calculate by direct integration the moment of inertia for a thin rod of mass \(M\) and length \(L\) about an axis located distance \(d\) from one end. Confirm that your answer agrees with Table 12.2 when \(d=0\) and when \(d=L / 2\).

An 8.0 -cm-diameter, \(400 \mathrm{g}\) sphere is released from rest at the top of a 2.1 -m-long, \(25^{\circ}\) incline. It rolls, without slipping, to the bottom. a. What is the sphere's angular velocity at the bottom of the incline? b. What fraction of its kinetic energy is rotational?

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