/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A bowler throws a bowling ball o... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(v_{\mathrm{com}, 0}=8.5 \mathrm{~m} / \mathrm{s}\) 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 \(v_{\text {com }}\) 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?

Short Answer

Expert verified
(a) \(v_{\text{com}} = \omega R\), (b) \(a = 2.058 \ \text{m/s}^2\), (c) \(\alpha = 46.91 \ \text{rad/s}^2\), (d) \(t \approx 1.81 \ \text{s} \), (e) \(\approx 12.7\ \text{m}\), (f) \(\approx 4.77\ \text{m/s} \).

Step by step solution

01

Relationship Between Linear and Angular Velocity

To stop sliding and roll smoothly, the condition is that the bowling ball's linear velocity \(v_{\text{com}}\) is equal to the product of its angular velocity \(\omega\) and its radius \(R\). Thus, the equation is \(v_{\text{com}} = \omega R\).
02

Determine Linear Acceleration

The force of kinetic friction, \( f_k = \mu_k m g \), is what decelerates the ball linearly. The linear acceleration is given by Newton's second law: \( a = \frac{f_k}{m} = \frac{\mu_k mg}{m} = \mu_k g \). Substituting \(\mu_k = 0.21\) and \(g = 9.8 \ \text{m/s}^2\), we have \( a = 0.21 \times 9.8 = 2.058 \ \text{m/s}^2\).
03

Determine Angular Acceleration

The torque due to friction is \( \tau = f_k R \), and this results in an angular acceleration \( \alpha \) given by \( \alpha = \frac{\tau}{I} \), where \( I = \frac{2}{5} m R^2 \) for a solid sphere. Therefore, \( \alpha = \frac{f_k R}{(2/5) m R^2} = \frac{5 f_k}{2 m R} \). Simplifying further, \( \alpha = \frac{5 \mu_k g}{2 R} \). Substituting, \( \alpha = \frac{5 \times 0.21 \times 9.8}{2 \times 0.11} \approx 46.91 \ \text{rad/s}^2\).
04

Determine Time of Sliding

During sliding, \(v_{\text{com}} - \omega R = 0\) must occur. The initial conditions give us \(v_{\text{com}},\omega_0=0\). Solving \(v_{\text{com}} - \omega R = 0\) : \( v_{\text{com}} = \omega R \Rightarrow a \cdot t = \alpha R \cdot t \). This implies : \( t = \frac{v_{\text{com}}}{a + \alpha R} = \frac{8.5}{2.058 + 46.91 \times 0.11} \approx 1.81 \ \text{s}\).
05

Calculate the Distance the Ball Slides

Using the formula for distance traveled, \(x = v_0 t + \frac{1}{2} a t^2\), with \(v_0 = 8.5\ \text{m/s}\), and substituting \(t = 1.81\ \text{s}\) and \(a = -2.058\ \text{m/s}^2\), we calculate:\[ x = 8.5 \times 1.81 + \frac{1}{2} \times (-2.058) \times (1.81)^2 \approx 12.7 \ \text{m}\].
06

Calculate Final Linear Speed

Using the relation \(v_f = v_0 + a t\), substitute \(v_0 = 8.5\ \text{m/s}\), \(a = -2.058\ \text{m/s}^2\), and \(t = 1.81\ \text{s}\), the final speed \(v_f\) can be found as:\[ v_f = 8.5 + (-2.058 \times 1.81) = 4.77 \ \text{m/s} \].

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Key Concepts

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

Linear Acceleration
Linear acceleration is a fundamental concept in physics, referring to the rate of change of velocity of an object moving in a straight line. In this context, the bowling ball experiences linear acceleration due to kinetic friction. The frictional force can be calculated using the formula \( f_k = \mu_k m g \), where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the ball, and \( g \) is the acceleration due to gravity, which is about \( 9.8 \, \text{m/s}^2 \). In this exercise, the coefficient of kinetic friction \( \mu_k \) is 0.21. Substituting these values gives us the linear acceleration \( a = \mu_k g = 0.21 \times 9.8 = 2.058 \, \text{m/s}^2 \). This negative acceleration indicates that the ball is decelerating due to the opposing force of friction acting on it as it moves forward.
Angular Acceleration
Angular acceleration represents the rate of change of angular velocity. It's crucial when dealing with rotational motion, such as a bowling ball spinning while sliding down the lane. Torque, which is the force causing rotational motion, is derived from the friction force; calculated by \( \tau = f_k R \). The moment of inertia \( I \) for a solid sphere is \( \frac{2}{5} m R^2 \). Using these, angular acceleration \( \alpha \) is given by the formula:\[ \alpha = \frac{\tau}{I} = \frac{f_k R}{(2/5) m R^2} = \frac{5 f_k}{2 m R} \].For this problem, substituting the known values \( \mu_k = 0.21 \), \( g = 9.8 \, \text{m/s}^2 \), and \( R = 0.11 \, \text{m} \) results in an angular acceleration \( \alpha = \frac{5 \times 0.21 \times 9.8}{2 \times 0.11} \approx 46.91 \, \text{rad/s}^2 \). This angular acceleration means the bowling ball's angular velocity increases as it slides, getting closer to the condition for smooth rolling.
Kinetic Friction
Kinetic friction plays a significant role in motion mechanics. Specifically, it is the force that opposes the sliding motion between two surfaces in contact. In the case of the bowling ball, it acts against the forward motion and provides the torque needed for angular acceleration.The coefficient of kinetic friction \( \mu_k \) is a dimensionless value that characterizes the interaction between the surfaces. Here, with \( \mu_k = 0.21 \), it indicates moderate friction between the ball and the lane. Kinetic friction is calculated using the formula \( f_k = \mu_k m g \), resulting in forces that contribute both to the ball's linear deceleration and its angular acceleration. Thus, kinetic friction is essential to understand the transition from sliding to smooth rolling — where the ball's forward speed equals the circumferential speed, fulfilling the condition \( v_{\text{com}} = \omega R \).This illustrates how kinetic friction does not merely slow objects down but also induces rotation, demonstrating the interconnectedness of linear and rotational dynamics.

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Most popular questions from this chapter

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