Problem 15 - Fundamentals of Physics

Question

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. \(11-38\) ) slides on the lane with initial speed \(v_{\mathrm{com}}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0 .\) The coefficient of kinetic friction between the ball and the lane is \(0.21 .\) The kinetic frictional force \(\vec{f}_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\operatorname{com}}\) in terms of \(\omega ?\) During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

Step-by-Step Solution

Answer

(a) \(v_{\text{com}} = 11\omega\), (b) \(-2.058 \, \text{m/s}^2\), (c) \(46.36 \, \text{rad/s}^2\), (d) \(0.178 \, \text{s}\), (e) \(1.421 \, \text{m}\), (f) \(0.9075\, \text{m/s}\).

Step 1: Determine the Relation between Linear and Angular Speed

To achieve smooth rolling, the condition is that the velocity of the center of mass \(v_{\text{com}}\) equals the product of angular speed \(\omega\) and radius \(R\). That relationship is given by: \[ v_{\text{com}} = \omega R \] Thus, \(v_{\text{com}} = 11 \omega\).

Step 2: Calculate Linear Acceleration

The frictional force \(f_k\) causes linear deceleration. It is given by:\[ f_k = \mu_k m g \]where \(\mu_k = 0.21\) is the friction coefficient, \(g = 9.8 \; \text{m/s}^2\) is the acceleration due to gravity, and \(m\) is the mass of the ball. The linear acceleration \( a \) (deceleration in this scenario) is:\[ a = -\frac{f_k}{m} = -\mu_k g \]Substituting the values, \( a = -0.21 \times 9.8 = -2.058 \, \text{m/s}^2 \).

Step 3: Calculate Angular Acceleration

The torque \(\tau\) caused by the frictional force provides angular acceleration. It is calculated by:\[ \tau = f_k R \]Using Newton’s second law for rotation, we find the angular acceleration \(\alpha\):\[ \alpha = \frac{\tau}{I} = \frac{\mu_k m g R}{\frac{2}{5}m R^2} \]Simplifying gives:\[ \alpha = \frac{5\mu_k g}{2R} \]Substitute \(\mu_k = 0.21\), \(g = 9.8\), and \(R = 0.11\, \text{m}\):\[ \alpha = \frac{5 \times 0.21 \times 9.8}{2 \times 0.11} \approx 46.36 \, \text{rad/s}^2 \].

Step 4: Calculate Time of Sliding

Initially, the angular speed \(\omega = 0\). Sliding stops when \(v_{\text{com}} = \omega R \). At that instant, we have:\[ v_{f} = v_{i} + a t \]and\[ \omega_{f} = \omega_{0} + \alpha t \]Equating \(v_{f}\) to \(\omega_{f} R\) and substituting the linear and angular acceleration equations:\[ 8.5 - 2.058t = 46.36t \times 0.11 \]\[ 8.5 = (46.36 \times 0.11 + 2.058)t \]\[ t = \frac{8.5}{47.7828} \approx 0.178 \, \text{s} \].

Step 5: Calculate Sliding Distance

The ball slides with decreasing velocity for the time found. The distance \(d\) can be determined by:\[ d = v_{i} t + \frac{1}{2} a t^2 \]Substitute \(v_i = 8.5\, \text{m/s}\), \(a = -2.058 \, \text{m/s}^2\) and \(t = 0.178\, \text{s}\):\[ d = 8.5 \times 0.178 + \frac{1}{2} (-2.058) \times 0.178^2 \approx 1.421 \, \text{m} \].

Step 6: Find Linear Speed for Smooth Rolling

Substitute back into \(v_{\text{com}} = \omega R \) using the calculation for \(t\):\[ \omega = \alpha \times t \approx 46.36 \times 0.178 \approx 8.25 \, \text{rad/s} \]\[ v_{\text{com}} = 8.25 \times 0.11 \approx 0.9075\, \text{m/s} \].

Key Concepts

Kinetic Friction

When a bowling ball slides down the lane, it encounters a force known as kinetic friction. This force acts opposite to the direction of motion and is responsible for both slowing down the linear movement of the ball and increasing its spin. The kinetic frictional force can be determined using the equation:

• \( f_k = \mu_k m g \)
\

The coefficient of kinetic friction, \( \mu_k \), is a constant value that represents how much friction there is between two surfaces. In this scenario, the coefficient is given as 0.21. The interplay between kinetic friction and the gravitational force, measured as \( g = 9.8 \, \text{m/s}^2 \), establishes the linear acceleration and angular acceleration experienced by the ball.
It’s important to understand that kinetic friction not only slows the ball down but also helps in the transition from sliding to rolling. This frictional force provides the necessary torque to increase the ball's angular velocity until it achieves a state where it rolls smoothly without slipping.

Linear Acceleration

Linear acceleration is the rate of change of velocity experienced by an object in motion. In the case of our bowling ball, kinetic friction is the sole factor that causes linear acceleration, specifically a form of deceleration since the ball is slowing down. The relationship can be described by:

• \( a = -\frac{f_k}{m} = -\mu_k g \)

The negative sign indicates that the force opposes the ball's motion. Substituting in our values, with \( \mu_k = 0.21 \) and \( g = 9.8\, \text{m/s}^2 \), we find the deceleration is: \[ a = -2.058 \, \text{m/s}^2 \].
This means that the bowling ball is slowing down at a rate of \( 2.058\, \text{m/s}^2 \) due to the frictional force. As the ball decelerates, it gradually moves towards achieving smooth rolling conditions.

Angular Acceleration

Angular acceleration is the rate at which the angular velocity of an object changes with time. In rolling motion, angular acceleration plays a crucial role in allowing a bowling ball to go from purely sliding to rolling without slipping. The frictional force provides a torque \( \tau \) that results in angular acceleration, calculated using:

• \( \alpha = \frac{5\mu_k g}{2R} \)

The formula incorporates the ball's radius \( R \), the kinetic friction coefficient \( 0.21 \), and the gravitational pull \( 9.8\, \text{m/s}^2 \). The resulting angular acceleration is: \[ \alpha \approx 46.36 \, \text{rad/s}^2 \].
This acceleration means that the ball's spin increases rapidly over time. As it spins faster, the angular acceleration helps the ball transition from sliding across the lane to smoothly rolling along it, where the linear speed of the ball matches its rotational speed times the radius.

Smooth Rolling Conditions

Smooth rolling conditions occur when a ball rolls down a lane without any slipping. This occurs when the linear velocity \( v_{\text{com}} \) of the ball's center of mass matches its angular velocity \( \omega \) times the radius \( R \). The condition is mathematically represented as:

• \( v_{\text{com}} = \omega R \)

During the transition period, while the ball is still sliding, both linear and angular accelerations affect it until the perfect balance or equivalence \( v_{\text{com}} = \omega R \) is achieved. Once this balance is reached, the sliding ceases.
For the bowling ball, this occurs when both angular and linear velocities align at the proper ratio, allowing the ball to roll smoothly. This marks the end of sliding and the onset of a steady rolling motion. This correlation between angular and linear parameters is essential in many rolling motion problems and highlights the harmonious nature required for perfect rolling conditions.