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A wheel rolls uniformly on level ground without slipping. A piece of mud on the wheel flies off when it is at the 9 o'clock position (rear of wheel). Describe the subsequent motion of the mud.

Short Answer

Expert verified
The mud follows a parabolic trajectory due to its initial horizontal velocity and gravity's influence.

Step by step solution

01

Understand the Initial Motion

The piece of mud is located on a wheel at the 9 o'clock position. This means that it is positioned horizontally towards the back of the wheel when it detaches. At the point of detachment, its velocity is purely horizontal in the direction of the wheel's motion, due to the wheel's rolling motion.
02

Analyze Velocity Components

When the mud leaves the wheel, its motion is determined by the velocity it had at the point of release. As the mud is on the wheel moving forward, it has a linear velocity from the wheel's forward motion as well as a tangential velocity from the wheel's rotation. Combining these gives a resultant velocity vector tangent to the path of rotation.
03

Apply Newton's First Law

Once the mud leaves the wheel, no forces act on it horizontally (assuming air resistance is negligible), so it continues to move in a straight line at a constant velocity according to Newton's First Law of Motion. This means the mud will follow a projectile motion path.
04

Determine the Path of the Mud

Initially, the mud moves horizontally forward, but as it continues in projectile motion under the influence of gravity, it follows a parabolic trajectory. The path of the mud will curve downwards in a parabola as it descends to the ground.

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Key Concepts

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

Newton's First Law
When discussing the motion of an object once it detaches from another, such as the piece of mud leaving the wheel, Newton's First Law is fundamental. This law states that an object in motion will remain in motion at a constant velocity unless acted upon by an external force.

In the context of the mud, once it detaches, its horizontal motion is unimpeded by additional forces (ignoring air resistance for simplicity). Because of this, the mud retains the velocity it had at the moment it was released. This means it will continue in a straight line in the direction of its velocity at the time it detached from the wheel. The absence of horizontal forces ensures this straight-line motion stays constant.

Understanding this principle helps explain why the mud follows a predictable path once it is no longer attached to the wheel, making it a key component in understanding projectile motion.
Velocity Components
The velocity of an object in motion can be broken down into components, typically horizontal and vertical. These components give insight into how an object moves through space. For the mud on the wheel, these different velocity components help determine its path after leaving the wheel.

  • **Horizontal Velocity Component**: This is derived from the forward motion of the wheel. As the wheel rolls, the point at which the mud is located moves horizontally with the wheel's progress.
  • **Tangential Velocity Component**: This arises due to the wheel's rotation. As the wheel rotates, each point on its edge moves tangentially relative to the wheel's center.

By combining these velocity components at the point of detachment, the mud begins its motion with a resultant velocity. This initial velocity vector is crucial in determining how the mud will travel through the air, laying the foundation for understanding its trajectory.
Rotational Motion
Rotational motion is involved whenever an object spins around a center point or axis. In our scenario, the wheel exhibits rotational motion as it rolls across the ground. This motion plays an essential role in the behavior of the mud when it's part of the wheel.

The piece of mud gains motion in two main ways due to rotation:
  • **Tangential Motion**: As the wheel spins, any point on its edge – including the mud – follows a curved path traveling around the wheel's circumference.
  • **Centripetal Force**: While attached to the wheel, the mud experiences a force pulling it towards the wheel's center. This force is only relevant while the mud is attached.

When the mud detaches at the 9 o'clock position, it leaves the rotational path and continues forward with the velocity of the wheel minus any centripetal influence. This transition is essential in setting the stage for projectile motion.
Parabolic Trajectory
A parabolic trajectory is the curved path an object follows under the influence of gravity. For the mud flying off the wheel, this trajectory begins with the initial velocity it had as part of the wheel's rotation.

Upon leaving the wheel, gravity progressively takes effect. The mud's horizontal component remains constant, but gravity introduces a vertical component that accelerates vertically downwards. The combined motion of these components gives rise to the characteristic parabolic shape.

Key points to understand the parabolic trajectory include:
  • **Influence of Gravity**: Causes the initially straight motion to curve downwards.
  • **Constant Horizontal Velocity**: As per Newton's First Law, unaffected by gravity.
  • **Shape of Path**: Forms a parabola that bends downward until reaching the ground.

The resulting path is predictable and can be described mathematically, offering a clear example of both projectile motion and the influence of gravity on an object in motion.

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Most popular questions from this chapter

A stationary ice skater with a mass of \(80.0 \mathrm{~kg}\) and a moment of inertia (about her central vertical axis) of \(3.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) catches a baseball with her outstretched arm. The catch is made at a distance of \(1.00 \mathrm{~m}\) from the central axis. The ball has a mass of \(145 \mathrm{~g}\) and is traveling at \(20.0 \mathrm{~m} / \mathrm{s}\) before the catch. (a) What linear speed does the system (skater \(+\) ball) have after the catch? (b) What is the angular speed of the system (skater \(+\) ball) after the catch? (c) What percentage of the ball's initial kinetic energy is lost during the catch? Neglect friction with the ice.

A uniform sphere and a uniform cylinder with the same mass and radius roll at the same velocity side by side on a level surface without slipping. If the sphere and the cylinder approach an inclined plane and roll up it without slipping, will they be at the same height on the plane when they come to a stop? If not, what will be the percentage difference of the heights?

A flat cylindrical grinding wheel is spinning at 2000 rpm (clockwise when viewed head-on) when its power is suddenly turned off. Normally, if left alone, it takes 45.0 s to coast to rest. Assume the grinder has a moment of inertia of \(2.43 \mathrm{~kg} \cdot \mathrm{m}^{2}\). (a) Determine its angular acceleration during this process. (b) Determine the tangential acceleration of a point on the grinding wheel if the wheel is \(7.5 \mathrm{~cm}\) in diameter. (c) The slowing down is caused by a frictional torque on the axle of the wheel. The axle is \(1.00 \mathrm{~cm}\) in diameter. Determine the frictional force on the axle. (d) How much work was done by friction on the system?

IE A small heavy object of mass \(m\) is attached to a thin string to make a simple pendulum whose length is \(L\) When the object is pulled aside by a horizontal force \(F\) it is in static equilibrium and the string makes a constant angle \(\theta\) from the vertical. (a) The tension in the string should be (1) the same as, (2) greater than, or (3) less than the object's weight, \(m g .\) (b) Use the force condition for static equilibrium (along with a free-body diagram of the object) to prove that the string tension is \(T=\frac{m g}{\cos \theta}>m g\). Use the same procedure to show that \(F=m g \tan \theta\). (c) Prove the same result for \(F\) as in part (b) using the torque condition, summing the torques about the string's tied end. Explain why you cannot use this method to determine the string tension.

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