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Chapter 10: Problem 30

A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle \(\beta\) to the horizontal (see Example 10.7 in Section 10.3 ). Treat the ball as a uniform solid sphere, ignoring the finger holes. (a) Draw the free-body diagram for the ball. Explain why the friction force must be directed uphill. (b) What is the acceleration of the center of mass of the ball? (c) What minimum coefficient of static friction is needed to prevent slipping?

Short Answer

Expert verified
The free-body diagram for the ball includes forces due to gravity, the normal force, and the friction force which points uphill. The acceleration of the center of mass for the ball can be calculated using its radius and the torque and moment of inertia; the minimum coefficient of static friction is determined from the static friction and normal forces and needs to balance the gravitational force component along the ramp to prevent slipping.

Step by step solution

01

Draw the free-body diagram

Start by drawing a diagram representing the object (in this case, the bowling ball) and all the forces that are acting on it. You should include: the gravitational force \(mg\) acting downward, the normal force \(N\) acting perpendicular to the surface, and the friction force \(f_s\) acting uphill.
02

Determine the direction of frictional force

Next, consider why the friction force is directed uphill. When the ball tries to roll up the slope, it has a tendency to slide down due to gravity. The frictional force opposes this potential sliding, and thus must be directed uphill.
03

Calculate the acceleration of the ball

To find the acceleration of the center of mass of the ball, use the equation for linear acceleration in rolling motion without slipping, which is \(a = r \cdot \alpha\), where \(r\) is the radius and \(\alpha\) is the angular acceleration. The angular acceleration \(\alpha\) can be obtained from the equation for rotational motion: \(\alpha = \frac{\tau}{I}\), where \(\tau\) is the net torque and \(I\) is the moment of inertia of the ball.
04

Find the minimum coefficient of static friction

Finally, to find the minimum coefficient of static friction necessary to prevent the ball from slipping, use the equation for static friction: \(f_s = \mu_s \cdot N\), where \(f_s\) is the static friction force, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force. To prevent slipping, the static friction force must be at least as large as the component of the gravitational force parallel to the surface of the ramp. So solve for \(\mu_s\), which gives \(\mu_s = \frac{f_s}{N}\).

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Key Concepts

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

Free-Body Diagram
Understanding physics problems often starts with a free-body diagram. It鈥檚 a simple yet powerful tool that visualizes all the forces acting on an object. For the rolling ball heading uphill, let's sketch the diagram.

A precise free-body diagram will show the ball at the center with three primary forces: the weight of the ball (gravitational force, denoted as \(mg\)), acting straight down towards the earth's center; the normal force (\(N\)), which is the surface鈥檚 push back on the ball, acting perpendicular to the slope; and the frictional force. It's crucial to depict the frictional force correctly, as it will indicate the direction of potential slip and the ball's actual movement, which, in this exercise, is uphill. This diagram helps us visualize how these forces interact and lay the foundation for our subsequent calculations and understanding of rolling motion physics.
Frictional Force Direction
The frictional force plays a crucial role when objects are in motion, especially on inclined planes. When we roll a ball uphill, gravity鈥檚 natural pull wants to drag it back down. Recognizing this, the frictional force must counteract gravity鈥檚 pull to prevent the ball from sliding backward. As a result, it's directed uphill supporting the ball's rolling motion up the ramp.

Here鈥檚 a simple way to comprehend this: If you try to push a book over a tilted desk, the book might slide down if not for the friction between the book and the desk. This frictional force always acts against the sliding motion - so in the case of our bowling ball rolling uphill, the frictional force must be upward along the slope, enabling the ball to roll without slipping. The correct direction of friction is vital in calculations and in practical applications like ensuring vehicle tires grip the road on a slope.
Coefficient of Static Friction
The coefficient of static friction (denoted as \(\mu_s\)) is a dimensionless number without units showing how strongly two surfaces cling to each other when at rest. It's pivotal for determining whether the ball can roll up the slope without slipping. The higher the coefficient, the greater the frictional forces needed to start moving the ball.

In our bowling ball scenario, to ensure the ball rolls without slipping, \(\mu_s\) has to be large enough so that the static friction force (\(f_s\)) overly matches the component of the gravitational force that's parallel to the slope. In mathematical terms, \(f_s \geq mg \sin(\beta)\), where \(\beta\) is the slope angle. This calculation informs us about the required texture or material of the ball and the ramp to achieve the no-slip condition, which is of immense value while designing materials for transport or sports equipment.

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Most popular questions from this chapter

A small \(10.0 \mathrm{~g}\) bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass \(50.0 \mathrm{~g}\) and is \(100 \mathrm{~cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of \(20.0 \mathrm{~cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

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.

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

A \(5.00 \mathrm{~kg}\) ball is dropped from a height of \(12.0 \mathrm{~m}\) above one end of a uniform bar that pivots at its center. The bar has mass 8.00 \(\mathrm{kg}\) and is \(4.00 \mathrm{~m}\) in length. At the other end of the bar sits another 5.00 \(\mathrm{kg}\) ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?
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