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A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Short Answer

Expert verified
The final angular speed of the rod is \(\frac{2v}{5L}\) and the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision is \(\frac{39}{50}\).

Step by step solution

01

Calculate Final Velocity

We know that the mass of the bullet is a fourth of the mass of the rod, so we can denote the mass of the bullet as \(m_b\) and the mass of the rod as \(4m_b\). The bullet embeds in the rod, so they will move together with the same velocity \(v'\) after collision. By applying the law of conservation of linear momentum (momentum before collision = momentum after collision), we get \(m_bv = (m_b + 4m_b)v'\). Solving for \(v'\) gives \(v'_ = \frac{v}{5}\).
02

Calculate Angular Speed

Now we find the angular velocity. We know that \(v = r\omega\), where \(r\) is the distance between collision point and pivot point. For our case, \(r = L/2\) because the bullet strikes the center of the rod. Substituting into the formula, we get \(\omega = \frac{v'}{r} = \frac{v}{5} \times 2/L = \frac{2v}{5L}\). This is the angular speed of the rod after the collision.
03

Calculate Kinetic Energy After Collision

The kinetic energy after the collision consists of two parts: the translational kinetic energy and the rotational kinetic energy. The total kinetic energy is the sum of these two, which is \[K.E = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2\], where \(I\) is the moment of inertia of the system about the pivot point. With a bit of calculation, we find that \(I = \frac{7L^2m}{12}\) in our case. Substituting and simplifying, we get \([K.E]_f = \frac{39v^2}{100}\).
04

Calculate Kinetic Energy Before Collision

The kinetic energy before the collision was only that of the bullet, \([K.E]_i = \frac{1}{2}mv^2\).
05

Find Ratio of Kinetic Energy

Finally, we are tasked with finding the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision. This is \[R =\frac{[K.E]_f}{[K.E]_i} = \frac{39v^2/(100)}{1/2v^2} = \frac{39}{50}\].

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Key Concepts

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

Conservation of Linear Momentum
Understanding the conservation of linear momentum is critical when analyzing collision events. Linear momentum, commonly just referred to as momentum, is a quantity that expresses the motion of an object and is defined as the product of an object's mass and its velocity.

In an isolated system where no external forces are acting, the total momentum before an event is equal to the total momentum afterward. This principle is known as the conservation of linear momentum. Applying this to collisions, such as the one involving the bullet and the rod in our exercise, ensures that we correctly account for how velocities change upon impact.

In a practical sense, for the bullet-rod collision, we assert that the momentum of the bullet before the collision is equal to the momentum of the bullet and rod combined after the collision. Since external forces (like friction or external fields) are absent, the bullet's momentum is transferred to the rod, causing both to move together. This concept not only helps determine the new velocity but also sets the stage for further calculations involving rotational motion.
Rotational Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. There are two types of kinetic energy relevant to our scenario: translational, for motion along a path, and rotational, for motion around an axis. Rotational kinetic energy is specifically the energy an object possesses due to its rotation and is given by the equation \( \frac{1}{2}I\omega^2 \) where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

In the case of the bullet embedding into the rod, we initially have translational kinetic energy of the bullet and none for the rod, as it is at rest. After the collision, the system acquires rotational kinetic energy because the rod (with the bullet embedded) starts to rotate about the pivot. This total energy after the collision is a sum of both the residual translational kinetic energy of the bullet and the newly gained rotational kinetic energy of the system. The exercise demonstrates how to calculate this sum and find the ratio of the kinetic energy post-collision to the kinetic energy pre-collision.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation rate. It plays a similar role in rotational motion as mass does in linear motion. The moment of inertia depends not only on the object's mass but also on how that mass is distributed relative to the axis of rotation.

In our textbook problem involving the rotating rod, we must calculate the moment of inertia to understand how the rod, combined with the bullet, will rotate after the collision. The moment of inertia can be quite complex to calculate, as it involves integrating the mass distribution over the object's volume. However, for simple shapes and mass distributions like a rod or a sphere, there are standard formulas one can use.

The calculated moment of inertia is then applied in the formula to find rotational kinetic energy, connecting it to the tangible outcome of how swiftly the rod turns post-collision. Knowing the correct moment of inertia is vital for accurately predicting how an object will behave when subjected to rotational forces, which is a fundamental aspect of dynamics in physics.

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Most popular questions from this chapter

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel \(0.650 \mathrm{~m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects \(0.200 \mathrm{~m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is \(8.00 \mathrm{~kg} ;\) its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at \(5.00 \mathrm{rev} / \mathrm{s}\). Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at \(0.300 \mathrm{rev} / \mathrm{s}\). (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle \(\beta\) to the horizontal (see Example 10.7 in Section 10.3 ). Treat the ball as a uniform solid sphere, ignoring the finger holes. (a) Draw the free-body diagram for the ball. Explain why the friction force must be directed uphill. (b) What is the acceleration of the center of mass of the ball? (c) What minimum coefficient of static friction is needed to prevent slipping?

In your job as a mechanical engineer you are designing a flywheel and clutch- plate system like the one in Example \(10.11 .\) Disk \(A\) is made of a lighter material than disk \(B\), and the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B .\) The moment of inertia of the shaft is negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0} ; B\) is initially at rest. The accelerating torque is then removed from \(A,\) and \(A\) is coupled to \(B\). (Ignore bearing friction.) The design specifications allow for a maximum of \(2400 \mathrm{~J}\) of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk \(A\) so as not to exceed the maximum allowed value of the thermal energy?

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to \(80.0 \mathrm{~N}\) is applied to the rim of the wheel. The wheel has radius \(0.120 \mathrm{~m}\) Starting from rest, the wheel has an angular speed of \(12.0 \mathrm{rev} / \mathrm{s}\) after \(2.00 \mathrm{~s}\). What is the moment of inertia of the wheel?

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