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Chapter 10: Problem 23

A solid ball is released from rest and slides down a hillside that slopes downward at \(65.0^{\circ}\) from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

Short Answer

Expert verified
The minimum coefficient of static friction between the hill and the ball for no slipping is given by \(\mu_s = \tan(65.0^{\circ})\). It is uncertain whether this would be sufficient for a hollow ball like a soccer ball without further information. The reason for using the static friction and not the kinetic is because we are considering a no slipping scenario where there is no relative motion at the point of contact.

Step by step solution

01

Calculate the forces acting on the solid ball

The forces acting on the solid ball include the gravitational force, normal force, and the frictional force. The gravitational force could be decomposed into a component parallel (\(F_{g \parallel}\)) and a component perpendicular (\(F_{g \perp}\)) to the inclined plane. Also, the static friction force \(\( f_s \)\) acts upward along the inclined plane. Mathematically, \(F_{g\parallel} = mg \sin(\theta)\), \(F_{g\perp} = mg \cos(\theta)\), and the maximum value of \(f_s\) can be expressed as \(\mu_s F_{g\perp}\).
02

Determine the condition for no slipping

For no slipping to occur, the frictional force should balance the component of the gravitational force that's along the incline. Therefore, \(f_s = F_{g\parallel}\). Substitute the equations from Step 1 into this equation and we can solve for \(\mu_s\). Hence, \(\mu_s = \tan(\theta)\).
03

Substitute the given value

Substitute \(\theta = 65.0^{\circ}\) into \(\mu_s = \tan(\theta)\) to get the minimum value of \(\mu_s\) required for no slipping to occur.
04

Discuss the applicability for a hollow ball

The hollow ball (like a soccer ball) would have a different distribution of mass and may require a different static friction coefficient to avoid slipping. However, we can't determine whether the calculated \(\mu_s\) is enough without more information.
05

Explain why static friction was used

In this context, static friction is used because we are analyzing a scenario under which the ball will not slip, i.e., there will be no relative movement between the surfaces at the point of contact. In such scenarios, static friction is applicable, not kinetic friction which requires relative motion.

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Key Concepts

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

Coefficient of Static Friction
The coefficient of static friction, denoted as \(\mu_s\), is a dimensionless number that represents the ratio of the maximum static frictional force between two surfaces to the normal force pushing them together. It is a critical value that indicates how much force is required to initiate motion between two objects at rest relative to each other.

For instance, in the case of a ball on an inclined plane, a higher \(\mu_s\) means more friction is available to prevent the ball from sliding down. The concept can be applied to everyday life, such as the grip of car tires on a road or the slip resistance of shoes on a floor. Mathematically, the maximum static frictional force can be expressed as \(f_s = \mu_s F_{g\perp}\), where \(F_{g\perp}\) is the perpendicular component of the gravitational force acting on the object.

Understanding the coefficient of static friction is essential in predicting whether an object will remain stationary or start moving when subjected to a force, which is particularly useful in engineering and safety calculations.
Forces on an Inclined Plane
When analyzing motions and forces on an inclined plane, recognizing and decomposing the relevant forces is fundamental. An object on a slope experiences gravity, which can be decomposed into two components: one parallel to the slope \(F_{g\parallel}\), which tries to move the object down the slope, and one perpendicular \(F_{g\perp}\), which is balanced by the normal force.

The parallel force is given by \(F_{g\parallel} = mg \sin(\theta)\) and is the force responsible for the likely movement of the object. The normal force and the perpendicular component of gravity counter each other why the normal force is calculated as \(F_{g\perp} = mg \cos(\theta)\). Frictional forces also play an important role on an incline, providing the resistance against the sliding motion. For an object to remain stationary, or for motion to begin, the balance between these forces needs to be understood.

To dive deeper, when calculating problems related to inclines, always start by clearly defining and resolving the forces acting on the object; this simplifies understanding and solving problems related to motion and stability on slopes.
Static Vs Kinetic Friction
The difference between static and kinetic friction is distinguishable by whether or not the objects in contact are moving relative to one another. Static friction occurs when the objects are stationary relative to each other \( - \) it is the force that must be overcome to start the motion of an object at rest. Kinetic friction, also known as sliding or dynamic friction, comes into play once motion has started.

Another noteworthy difference is the magnitude: static friction is generally greater than kinetic friction for the same pair of materials. This means it takes more force to initiate the movement than to maintain it. The coefficient of kinetic friction \(\mu_k\) is typically used to calculate the force of friction once an object has started moving.

In the provided exercise, static friction is relevant because the ball starts from rest, and we want to know if it will begin to slide down the inclined plane. If there were motion involved, we would switch our focus to kinetic friction to understand how the motion would be affected after it had started.

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Most popular questions from this chapter

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

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 block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

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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