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Chapter 16: Problem 94

A wheel of radius \(r\) and centroidal radius of gyration \(\bar{k}\) is released from rest on the incline and rolls without sliding. Derive an expression for the acceleration of the center of the wheel in terms of \(r, \bar{k}, \beta,\) and \(g .\)

Short Answer

Expert verified
The acceleration is \( a = \frac{g \sin(\beta)}{1 + \left(\frac{\bar{k}}{r}\right)^2} \).

Step by step solution

01

Analyze Forces Acting on the Wheel

Consider a wheel of radius \(r\) and mass \(m\) rolling down an incline with angle \(\beta\). The forces acting on the wheel are gravity \(mg\), normal force \(N\), and frictional force \(f\). Since the wheel rolls without sliding, friction acts only to prevent slipping and does no work.
02

Apply Newton's Second Law

We apply Newton's Second Law for translational motion along the incline:\[ mg \sin(\beta) - f = ma \]where \(a\) is the linear acceleration of the center of the wheel.
03

Consider Rotational Motion and Moment of Inertia

For rotational motion about the center of mass, which is given as:\[ \text{Moment of Inertia} = I = m\bar{k}^2 \].The wheel's angular acceleration \(\alpha\) is related to its linear acceleration by \(a = r\alpha\).
04

Apply Newton's Second Law for Rotation

For rotational motion, apply Newton鈥檚 second law:\[ f \cdot r = I\alpha = m\bar{k}^2\frac{a}{r} \]Solve for \(f\):\[ f = \frac{m\bar{k}^2 a}{r^2} \]
05

Solve for Linear Acceleration \(a\)

Substitute the expression for \(f\) from the rotational analysis into the translational equation:\[ mg \sin(\beta) - \frac{m\bar{k}^2 a}{r^2} = ma \]Rearrange and solve for \(a\):\[ a = \frac{g \sin(\beta)}{1 + \frac{\bar{k}^2}{r^2}} \]
06

Derive Final Expression

Conclude with the derived expression for the acceleration:\[ a = \frac{g \sin(\beta)}{1 + \left(\frac{\bar{k}}{r}\right)^2} \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a property of rotating objects that measures how difficult it is to change their rotational motion. It is similar to mass in linear motion. In this problem, the wheel's moment of inertia helps us understand how its mass is distributed relative to the axis of rotation.
The moment of inertia is given by the formula: - \[ I = m\bar{k}^2 \] where:
  • \( I \) is the moment of inertia.
  • \( m \) is the mass of the wheel.
  • \( \bar{k} \) is the radius of gyration.
Since the wheel is rolling without slipping, the moment of inertia plays a crucial role in determining the relationship between the forces acting and the resulting motion. It accounts for how the mass of the wheel is distributed around its center, impacting both its linear and rotational dynamics.
Newton's Second Law
Newton's Second Law is fundamental in explaining the motion of objects. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. For the wheel rolling down the incline:- \[ F = ma \] When we consider the forces acting along the incline, the equation becomes: - \[ mg \sin(\beta) - f = ma \] where:
  • \( m \) is the mass of the wheel.
  • \( g \) is the acceleration due to gravity.
  • \( \beta \) is the angle of incline.
  • \( f \) is the frictional force.
  • \( a \) is the linear acceleration.
This equation helps us establish the linear relationship between forces and acceleration. We use it alongside rotational dynamics to derive expressions for acceleration.
Rotational Motion
Rotational motion describes how objects spin around an axis. For a rolling wheel, rotational motion is linked to the linear motion at its center. The crucial relationship here is - \[ a = r \alpha \] where:
  • \( a \) is the linear acceleration.
  • \( r \) is the radius of the wheel.
  • \( \alpha \) is the angular acceleration.
Rotational motion for the wheel combines forces into spinning, linking angular acceleration \( \alpha \) with the linear acceleration \( a \). Through rotational dynamics principles, such as torque represented by friction, we can use the rotational form of Newton's second law: \[ f \cdot r = I \alpha \] This equation connects friction and moment of inertia, establishing how these influence the wheel's motion.
Linear Acceleration
Linear acceleration refers to how quickly an object's velocity changes as it moves along a straight path. By examining both translational and rotational dynamics, we can find the linear acceleration of the wheel's center of mass.
Using the combination of Newton's second law for both linear and rotational aspects, we derive:- \[ a = \frac{g \sin(\beta)}{1 + \left(\frac{\bar{k}}{r}\right)^2} \] This equation synthesizes all the elements:
  • The gravitational pull along the incline \( mg \sin(\beta) \).
  • The moment of inertia's impact through \( \left(\frac{\bar{k}}{r}\right)^2 \).
  • The resistance due to rotational inertia.
Through this derived formula, we see how various forces and wheel properties interact dynamically to determine the speed at which the wheel accelerates down the incline.

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Most popular questions from this chapter

Disk \(A\) has a mass of \(6 \mathrm{kg}\) and an initial angular velocity of \(360 \mathrm{rpm}\) clockwise; disk \(B\) hass of \(3 \mathrm{kg}\) and is initially at rest. The disks are brought together by applying a horizontal force of magnitude \(20 \mathrm{N}\) to the axle of disk \(A .\) Knowing that \(\mu_{k}=0.15\) between the disks and neglecting bearing friction, determine \((a)\) the angular acceleration of each disk, \((b)\) the final angular velocity of each disk.

The rotor of an electric motor has an angular velocity of \(3600 \mathrm{rpm}\) when the load and power are cut off. The \(120-1 b\) rotor, which has a centroidal radius of gyration of \(9 \mathrm{in}\), then coasts to rest. Knowing that kinetic friction results in a couple of magnitude 2.5 lb. ft exerted on the rotor, determine the number of revolutions that the rotor executes before coming to rest.

The motion of the uniform slender rod of length \(L=0.5 \mathrm{m}\) and mass \(m=3 \mathrm{kg}\) is guided by pins at \(A\) and \(B\) that slide freely in frictionless slots, circular and horizontal, cut into a vertical plate as shown. Knowing that at the instant shown the rod has an angular velocity of 3 rad/s counterclockwise and \(\theta=30^{\circ},\) determine the reactions at points \(A\) and \(B .\)

A 15 -ft beam weighing 500 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration of cable \(A\) is \(20 \mathrm{ft} / \mathrm{s}^{2}\) and the deceleration of cable \(B\) is \(2 \mathrm{ft} / \mathrm{s}^{2}\), determine the tension in each cable.

A uniform slender rod of length \(L=900 \mathrm{mm}\) and mass \(m=4 \mathrm{kg}\) is suspended from a hinge at \(C .\) A horizontal force \(\mathrm{P}\) of magnitude \(75 \mathrm{N}\) is applied at end \(B\). Knowing that \(\bar{r}=225 \mathrm{mm}\), determine \((a)\) the angular acceleration of the rod, \((b)\) the components of the reaction at \(C\).
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