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A small rubber ball of radius \(r\) is thrown against a rough floor with a velocity \(\overline{\mathbf{v}}_{A}\) of magnitude \(\mathbf{v}_{0}\) and a backspin \(\omega_{A}\) of magnitude \(\omega_{0}\). It is observed that the ball bounces from \(A\) to \(B\), then from \(B\) to \(A\), then from \(A\) to \(B,\) etc. Assuming perfectly elastic impact, determine the required magnitude \(\omega_{0}\) of the backspin in terms of \(\bar{v}_{0}\) and \(r .\)

Short Answer

Expert verified
The required backspin magnitude \( \omega_{0} = \frac{\bar{v}_{0}}{r} \).

Step by step solution

01

Understanding Elastic Impact

In a perfectly elastic impact, the relative velocities between the ball and the floor remain constant but reverse in direction. The ball's velocity component parallel to the floor changes from \ \( \bar{v}_{A} \ \) to \ \( -\bar{v}_{A} \ \), and the same applies to the rotational component due to backspin.
02

Velocity and Spin Relation at Landing

When the ball strikes the floor, its horizontal component of translational velocity is \ \( \bar{v}_{0} \ \) and rotational velocity due to backspin is \ \( \omega_{A} \times r \ \). For the ball not to slip, the condition \ \( \bar{v}_{0} = \omega_{0} \times r \ \) must be satisfied.
03

Solving for Backspin

Using the non-slip condition, rearrange the equation \ \( \bar{v}_{0} = \omega_{0} \times r \ \) to find the backspin in terms of translational velocity and radius, i.e., \ \( \omega_{0} = \frac{\bar{v}_{0}}{r} \ \).

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Key Concepts

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

Backspin Dynamics
Backspin dynamics refers to the motion of a rotating object, such as a ball, where it spins opposite to its forward motion. When a rubber ball is thrown with backspin, the bottom part of the ball moves backward relative to its translational direction. As a result, this spin affects how the ball will behave when it impacts a surface.

For a ball with backspin hitting a rough floor, the rotational direction is initially opposite to the translational direction. This impacts the interaction with the floor because the spin exerts a torque that can alter the ball's motion. The magnitude of this spin is denoted as \( \omega_0\), and it plays a significant role in the resulting bounce dynamics. Think of backspin as giving the ball more control over its trajectory through the bounce.

In sports, backspin is often used deliberately to modify a ball's path or its behavior upon contact with surfaces, such as ensuring consistent bounces or reducing forward motion tendency. Mastering backspin dynamics is crucial for understanding how to control a ball in various practical scenarios.
Non-slip Condition
In the context of backspin, the non-slip condition is essential to determining how a ball interacts with the floor upon impact. This condition ensures that there's no slipping between the ball's surface and the floor, implying that the point of contact stays stationary relative to the floor at the moment of impact.

The non-slip condition is mathematically defined as \( \ar{v}_{0} = \omega_{0} \times r \), where \( \ar{v}_{0}\) is the horizontal velocity, \( \omega_{0}\) is the angular velocity (backspin), and \( r\) is the ball's radius. What this equation tells us is that the translational speed at which the point of contact tends to move along the floor should equal the rotational speed caused by the backspin.

Under this condition, the frictional force that usually causes slipping is neutralized, allowing the ball to roll smoothly at the point of contact. This is particularly important in designing situations where precision of the ball's path and bounce is needed, such as in sports like table tennis or golf.
Translational and Rotational Motion
Understanding translational and rotational motion helps in dissecting the complex movements of objects like a spinning ball. Translational motion refers to the movement in which all parts of the ball move in the same linear direction. For a ball tossed onto a floor, this is captured by its velocity \( \ar{v}_{0}\).

Rotational motion, on the other hand, describes the spin of the ball around its axis. This is where the backspin \( \omega_{0}\) comes into play. The ball does not merely travel in a straight line but also spins around its center, adding dynamism to its motion.
  • Translational Motion: Direct linear movement defined by velocity vector \( \ar{v}_{0}\).
  • Rotational Motion: Angular movement around an internal axis, characterized by angular velocity \( \omega_{0}\).
When analyzing motion dynamics, combining these two forms helps predict the overall behavior of the ball. For example, a perfectly elastic bounce involves both reflecting the speed and reversing the direction of these motions. Consider both when examining ball trajectories in games or any scenario involving bouncing objects.

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Most popular questions from this chapter

The flywheel of a small punch rotates at 300 rpm. It is known that 1800 ft?lb of work must be done each time a hole is punched. It is desired that the speed of the flywheel after one punching be not less than 90 percent of the original speed of 300 rpm. (a) Determine the required moment of inertia of the flywheel. (b) If a constant 25-lb?ft couple is applied to the shaft of the flywheel, determine the number of revolutions that must occur between each punching, knowing that the initial velocity is to be 300 rpm at the start of each punching.

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