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Chapter 7: Problem 28

The solid right-circular cone of base radius \(r\) and height \(h\) rolls on a flat surface without slipping. The center \(B\) of the circular base moves in a circular path around the \(z\) -axis with a constant speed \(v\) Determine the angular velocity \(\omega\) and the angular acceleration \(\alpha\) of the solid cone.

Short Answer

Expert verified
The angular velocity \( \omega = \frac{v}{\sqrt{r^2 + h^2}} \) and the angular acceleration \( \alpha = 0 \).

Step by step solution

01

Understand the problem

We need to find the angular velocity \( \omega \) and angular acceleration \( \alpha \) of a right-circular cone while it's rolling around the \( z \)-axis.
02

Relate linear and angular speed

Since the cone rolls without slipping, the relationship between linear speed \( v \) and angular speed \( \omega \) is given by \( v = R\omega \), where \( R \) is the distance from the axis of rotation \( z \) to the center \( B \) of the cone's base.
03

Determine the radius of the path

The radius \( R \) of the path that \( B \) follows is equivalent to the slant height of the cone's lateral surface, which can be found using the Pythagorean theorem: \[ R = \sqrt{r^2 + h^2} \].
04

Calculate the angular velocity \( \omega \)

Using the relation from Step 2, \( \omega = \frac{v}{R} \). Substitute \( R \) from Step 3 to get: \[ \omega = \frac{v}{\sqrt{r^2 + h^2}} \].
05

Determine angular acceleration \( \alpha \)

Since the problem specifies a constant speed \( v \) for \( B \), \( \omega \) also remains constant. Therefore, the angular acceleration \( \alpha = \frac{d\omega}{dt} = 0 \).

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Key Concepts

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

Rolling Motion
Rolling motion is a fascinating type of motion combining both linear and rotational movements. It occurs when an object moves on a surface in such a way that it rolls rather than slides or slips. In the problem of a right-circular cone, the cone rolls on a flat surface, meaning its base maintains contact without slipping.

This type of motion is present in many day-to-day scenarios like:
  • A bicycle wheel moving on a road.
  • A ball rolling across a floor.
  • Even a cone rolling as it describes a circular path.
In our case, since the cone rolls without slipping, there exists a direct relationship between its linear speed and angular velocity. This configuration ensures that every point on the base of the cone makes contact with a different point on the surface progressively, while all these points on its base move at the same linear speed at every instant. This relationship is crucial when determining both the angular velocity and the angular acceleration for the cone.
Circular Motion
Circular motion is a type of movement where an object travels around a fixed point along a circular path. The movement around this point is characterized by a central axis, around which the entire motion takes place.

In the present problem, the center of the cone's base moves in a circular path around the z-axis. This motion is particularly interesting because it involves the following elements:
  • Radius of the path: This is the distance from the z-axis to the center of the cone's base. In our scenario, it is equivalent to the slant height of the cone and is found using the equation: \[ R = \sqrt{r^2 + h^2} \]
  • Angular velocity (\omega): It is the rate of change of the angle through which the radius vector pivots. For the cone, it can be calculated using \omega = \frac{v}{R}\ .
  • Angular acceleration (\alpha): Describes how the angular velocity changes with time; however, in this problem, this value is zero since the speed is constant.
The analysis of circular motion assists in examining rolling motion's impact, helping to calculate the angular properties of the object in question.
Right-Circular Cone
The right-circular cone in this problem is a three-dimensional geometric shape characterized by a circular base and a vertex point, which lies above or below the center of the base. It has certain specific properties that are essential for understanding its motion.Here are some of those properties:
  • Base radius ( r ): This is the radius of the circular base of the cone.
  • Height ( h ): The perpendicular distance from the base to the cone's vertex.
  • Slant height: This is the distance along the cone's lateral side from the base to the vertex. It can be calculated using Pythagoras' theorem: \[ R = \sqrt{r^2 + h^2} \].
The function of the cone in this problem is not just limited to its geometric dimensions. Its motion, characterized by its rolling and movement along a circular path, is heavily reliant on these dimensions to calculate the angular velocity and determine the other aspects of its kinematic behavior.

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Most popular questions from this chapter

The flight simulator is mounted on six hydraulic actuators connected in pairs to their attachment points on the underside of the simulator. By programming the actions of the actuators, a variety of flight conditions can be simulated with translational and rotational displacements through a limited range of motion. Axes \(x-y-z\) are attached to the simulator with origin \(B\) at the center of the volume. For the instant represented, \(B\) has a velocity and an acceleration in the horizontal \(y\) -direction of \(3.2 \mathrm{ft} / \mathrm{sec}\) and \(4 \mathrm{ft} / \mathrm{sec}^{2},\) respectively. Simultaneously, the angular velocities and their time rates of change are \(\omega_{x}=1.4 \mathrm{rad} / \mathrm{sec}, \dot{\omega}_{x}=2 \mathrm{rad} / \mathrm{sec}^{2}, \omega_{y}=1.2 \mathrm{rad} / \mathrm{sec}\) \(\dot{\omega}_{y}=3 \mathrm{rad} / \mathrm{sec}^{2}, \omega_{z}=\dot{\omega}_{z}=0 .\) For this instant determine the magnitudes of the velocity and acceleration of point \(A\)

A dynamics instructor demonstrates gyroscopic principles to his students. He suspends a rapidly spinning wheel with a string attached to one end of its horizontal axle. Describe the precession motion of the wheel.

The 8 -lb rotor with radius of gyration of 3 in. rotates on ball bearings at a speed of 3000 rev/min about its shaft \(O G\). The shaft is free to pivot about the \(X\) -axis, as well as to rotate about the \(Z\) -axis. Calculate the vector \(\Omega\) for precession about the \(Z\) -axis. Neglect the mass of shaft \(O G\) and compute the gyroscopic couple M exerted by the shaft on the rotor at \(G\)

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The housing of the electric motor is freely pivoted about the horizontal \(x\) -axis, which passes through the mass center \(G\) of the rotor. If the motor is turning at the constant rate \(\dot{\phi}=p,\) determine the angular acceleration \(\ddot{\psi}\) which will result from the application of the moment \(M\) about the vertical shaft if \(\dot{\gamma}=\dot{\psi}=0 .\) The mass of the frame and housing is considered negligible compared with the mass \(m\) of the rotor. The radius of gyration of the rotor about the \(z\) -axis is \(k_{z}\) and that about the \(x\) -axis is \(k_{x}\)
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