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Chapter 10: Problem 24

A hollow, spherical shell with mass \(2.00 \mathrm{~kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to \(4.00 \mathrm{~kg}\) ?

Short Answer

Expert verified
The acceleration \( a \) and the frictional force \( f \) can be calculated by the formulas given above. The value of \( \mu \) would be \( \mu \geq \frac{2f_{max}}{3mg \cos(\theta)} \). When the mass is doubled, \( a \) and \( f \) double, while \( \mu \) remains the same.

Step by step solution

01

Calculate the gravitational force on the shell

Due to the gravitational pull of the earth, there exists a force \( F_{g} = mg \) acting downwards on the spherical shell, where \( m = 2.00 \, \mathrm{kg} \) is the mass of the shell and \( g = 9.8 \, \mathrm{m/s}^{2} \) is the acceleration due to gravity. However, only the component of this force along the slope contributes to moving the shell, which is \( F_{gs} = mg \sin(\theta) \), with \( \theta = 38.0^{\circ} \) being the angle of the slope.
02

Calculate the equation of motion

The shell rolls without slipping, which means we can use the equation of rotational motion: \( F = ma \) for linear motion and \( I\alpha = \tau \) for rotational motion, where \( I \) is the moment of inertia of the shell, \( \alpha \) is the angular acceleration, and \( \tau \) is the torque. Given that the shell is hollow, its moment of inertia \( I = \frac{2}{3}mr^{2} \). Since \( \alpha = a/r \), the torque \( \tau = I\alpha = \frac{2}{3}mra \). Together with the pictorial consideration that the frictional force equals the torque (that is, \( f = \tau \)), we obtain the equation of motion \( mg \sin(\theta) - f = ma \), where \( f \) is the frictional force.
03

Solve for \( a \)

In the equation of motion obtained in step 2, solve for \( a \), which yields \( a = \frac{3gs \sin(\theta)}{2 + 3 \mu} \), where \( \mu \) is the coefficient of static friction.
04

Solve for frictional force

Using the relationship from step 2 that \( f = \frac{2}{3}mra \), we can use the \( a \) calculated in the previous step to find \( f \). Additionally, we know that \( f \leq \mu N \), where \( N = mg \cos(\theta) \) is the normal force.
05

Repeat for doubled mass

Repeat the calculations from steps 2 to 4 but with a doubled mass \( m = 4.00 \, \mathrm{kg} \), and compare the results with those from the previous steps.

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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 key concept in rotational dynamics. It measures how difficult it is to change an object's rotational speed. Think of it like the rotational equivalent of mass in linear motion.

For a hollow spherical shell, the moment of inertia is given by the formula:
  • \( I = \frac{2}{3}mr^{2} \)
Here, \( m \) is the mass of the shell and \( r \) is its radius. The moment of inertia depends heavily on how mass is distributed relative to the axis of rotation:
  • The further the mass is from the axis, the larger the moment of inertia.
  • Shapes like hollow spheres have unique inertia formulas reflecting their mass distribution.
Understanding moment of inertia helps us calculate how forces affect an object's rotational motion.
Rotational Dynamics
Rotational dynamics deals with forces and motions that cause an object to rotate. It's like using Newton's second law for spinning objects.

Key equations in rotational dynamics include:
  • \( F = ma \) for linear motion.
  • \( \tau = I\alpha \) for rotational motion, where \( \tau \) is torque, and \( \alpha \) is angular acceleration.
Torque is essentially a force that causes rotation, much like how pushing a door causes it to swing open. In our scenario, the shell's rotational motion is influenced by the forces on it:
  • Torque is produced by the frictional force \( f = \tau \).
  • The angular acceleration \( \alpha \) relates to linear acceleration \( a \) by \( \alpha = \frac{a}{r} \).
Understanding these principles is crucial for analyzing how objects rotate under various forces.
Static Friction
Static friction is the force that keeps an object at rest when a force tries to move it. It's the reason our shell can roll without slipping down the slope.

For an object rolling without slipping:
  • The frictional force \( f \) provides the necessary torque for rotation.
  • It must be less than or equal to the maximum static friction, \( f \leq \mu N \), where \( \mu \) is the coefficient of static friction and \( N \) is the normal force.
The static frictional force needs to be just right:
  • If too low, the shell might slip.
  • If too high, it can hinder movement.
This balance allows the shell to roll smoothly down the incline, maintaining the rotational motion necessary to avoid slipping.
Gravitational Force
Gravitational force is the force of attraction between two masses. It's why objects fall to the ground when dropped. Here, it acts on the shell, pulling it down the slope.

For our problem, we focus on the component of gravitational force along the slope:
  • Given by \( F_{gs} = mg \sin(\theta) \), where \( \theta \) is the slope angle.
  • This component propels the shell downwards.
Gravitational force also acts perpendicular to the slope:
  • This component, \( mg \cos(\theta) \), determines the normal force \( N \).
In effect, gravity's role is to provide the energy that drives the shell's movement and affects the frictional requirements for rolling.

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Most popular questions from this chapter

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel \(0.650 \mathrm{~m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects \(0.200 \mathrm{~m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is \(8.00 \mathrm{~kg} ;\) its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at \(5.00 \mathrm{rev} / \mathrm{s}\). Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at \(0.300 \mathrm{rev} / \mathrm{s}\). (d) At what rate must the shaft rotate in order that it may be supported at one end only?
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