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Chapter 11: Problem 4

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of \(0.10 \mathrm{~g} ?\) (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to \(0.10 \mathrm{~g}\) ? Why?

Short Answer

Expert verified
The incline angle should be approximately \(8.05^\circ\). A frictionless block would have a greater acceleration of \(0.14g\).

Step by step solution

01

Identify the Forces Involved

The forces acting on the sphere include gravitational force, normal force, and friction. As the sphere rolls down the incline, it benefits from the torque due to friction, allowing it to roll without slipping.
02

Use Newton's Second Law

For a sphere rolling down the incline without slipping, both translational and rotational dynamics are involved. We apply Newton's second law for translation as\[ m a = mg \sin \theta - f \]where \( f \) is friction, \( m \) is mass, \( g \) is acceleration due to gravity, \( a \) is linear acceleration, and \( \theta \) is the incline angle.
03

Apply Rotational Dynamics

For rotation,\[ I \alpha = f r \]where \( I \) is the moment of inertia \( \left(I = \frac{2}{5}mr^2\right)\), \( \alpha \) is angular acceleration, and \( r \) is the sphere's radius. The relationship between \( a \) and \( \alpha \) for rolling is \( a = r \alpha \). Simplify to find \( f \) in terms of \( m, r, \) and \( a \).
04

Solve for Friction and Substitute

Solving both translational and rotational equations simultaneously gives us:\[ f = \frac{2}{7}mg \sin \theta \]Substitute this back into the translational equation to express \( a \):\[ a = \frac{5}{7} g \sin \theta \]
05

Determining Angle

To find the angle such that \( a = 0.1g \), use:\[ 0.1g = \frac{5}{7}g \sin \theta \]This simplifies to:\[ \sin \theta = \frac{0.1 \times 7}{5} = 0.14 \]Thus, \( \theta = \arcsin(0.14) \approx 8.05^\circ \).
06

Consider the Frictionless Block

For a frictionless block sliding down the same incline angle, the acceleration is given by:\[ a = g \sin \theta \]Given that \( \sin \theta = 0.14 \), the block's acceleration is \( 0.14g \), which is greater than \(0.10g\).

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Key Concepts

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

Understanding Incline Angle
When an object rests on an incline, the inclination angle \(\theta\) is the angle between the inclined surface and the horizontal plane. This angle plays a vital role in determining how objects move down the slope. For a rolling sphere, the incline angle affects both translational and rotational movements.
  • Higher incline angles result in a steeper slope and typically, faster acceleration.
  • When solving problems involving a rolling sphere or block on a slope, \(\sin(\theta)\) is often used to express how the gravitational component along the incline interacts with the object.
In the original problem, we use the sine of the incline angle to equate the forces involved through Newton's laws to find the exact angle needed for a specific linear acceleration. Understanding how to manipulate this angle helps solve motion equations effectively.
Exploring Linear Acceleration
Linear acceleration refers to the rate of change of velocity of an object moving in a straight path. When discussing rolling motion, like a sphere on an incline, linear acceleration determines how quickly the sphere's center of mass speeds up.Consider these key points:
  • Linear acceleration \(a\) occurs due to gravity's pull down the incline, described by the equation \(a = \frac{5}{7}g \sin \theta\), where \(g\) is gravitational acceleration.
  • The sphere's linear acceleration is less than that of a sliding block because some energy is used in rotational motion due to friction.
The problem asks for the incline angle needed for a linear acceleration magnitude of \(0.1g\). By relating \(a\) to \(g\sin \theta\), we isolated \(\theta\) to be approximately \(8.05^\circ\). This solution highlights how linear motion is interconnected with rotational dynamics.
Role of Friction in Rolling Motion
Friction is essential for rolling without slipping. It provides the necessary torque for rotational motion and keeps the object from merely sliding. In rolling motion problems, friction acts parallel to the contact surface, preventing sliding and enforcing rolling.Let's break down the basics:
  • Frictional force \(f\) is involved in generating rotational acceleration in the sphere: \(f = \frac{2}{7}mg \sin \theta\).
  • This force ensures that all points of the sphere do not slide but roll smoothly, converting some gravitational energy into rotational motion.
Interestingly, if friction were absent (imagine a frictionless block), the block would experience higher linear acceleration because there would be no rotational energy loss鈥攎ore of the gravitational force would convert directly into linear motion. This explains why friction substantially affects both linear and rotational outcomes on an incline.

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Most popular questions from this chapter

A Texas cockroach of mass \(0.17 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(15 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2},\) and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s},\) and the lazy Susan turns clockwise with angular speed \(\omega_{0}=2.8 \mathrm{rad} / \mathrm{s} .\) The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

At one instant, force \(\vec{F}=4.0 \mathrm{j} \mathrm{N}\) acts on a \(0.25 \mathrm{~kg}\) object that has position vector \(\vec{r}=(2.0 \mathrm{i}-2.0 \mathrm{k}) \mathrm{m}\) and velocity vector \(\vec{v}=(-5.0 \hat{i}+5.0 \mathrm{k}) \mathrm{m} / \mathrm{s}\). About the origin and in unit-vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m},\) its inner radius \(R_{1}\) is \(R_{2} / 2.00\) its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

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