A Mass-Spring-Damper Model of a Bouncing Ball M. Nagurka and S. Huang Department of Mechanical and Industrial Engineering Marquette University, Milwaukee, WI 53201, USA [email protected], [email protected] Abstract— The mechanical properties of a vertically dropped ball, represented by an equivalent mass-spring- damper model, are related to the coefficient of restitution Impact and the time of contact of the ball during one bounce with the impacting surface. In addition, it is shown that Deformation Restitution the coefficient of restitution and contact time of a single x bounce are related to the total number of bounces and the h 0 h total time elapsing between dropping the ball and the ball 1 coming to rest. For a ball with significant bounce, approximate expressions for model parameters, i.e., stiffness and damping or equivalently natural frequency and damping ratio, are developed. Experimentally-based results for a bouncing ping- Initial Initial Maximum Final Maximum pong ball are presented. Condition Contact Deflection Contact Rebound (Dropped (After First From Rest) Bounce) I. INTRODUCTION The bouncing behavior of a dropped ball is a classic Fig. 1. A mass-spring-damper model of a ball showing phases in impact problem studied in depth [1]-[5]. The topic is treated in at first bounce. virtually all textbooks of physics and dynamics that address the subject of impact. These books also present, but in a separate section, the concept of mass, stiffness, and damping initial conditions are x(0) = h0 and x_(0) = 0 for a ball as the three elemental properties of a mechanical system. released from rest from height h0. The solution of this sim- To the authors’ knowledge, the textbooks and references do ple problem appears in physics and mechanics textbooks, not make a connection between the mechanical “primitives” leading to the classical results of vertical projectile motion. of mass, stiffness and damping and the coefficient of resti- When the ball is in contact with the ground, deformation tution, presented as part of the subject of impact. This paper and restitution occur. The equation of motion is then, develops this connection for a particular system, namely a mxÄ + cx_ + kx = ¡mg (2) bouncing ball, represented by a linear mass-spring-damper model. It is shown that the properties of the ball model with the initial conditions of x(0) = 0 and x_(0) = ¡v0 can be related to the coefficient of restitution and bounce where v0 is the velocity of the ball just prior to contact contact time. Furthermore, for the vertically dropped ball with the ground. Integrating eq. (2) gives problem it is shown that the total number of bounces and the · ¸ cg ¡ 2kv mg total bounce time, two parameters that are readily available x = 0 sin ! t + cos ! t £ 2k! d k d experimentally, can be related to the stiffness and damping. d ³ c ´ mg The analytical findings are tested to predict model properties exp ¡ t ¡ (3) 2m k of a ping-pong ball. where the damped natural frequency, !d, is II. MASS-SPRING-DAMPER MODEL 1 p ! = 4km ¡ c2 : (4) To study the behavior of a vertically dropped ball, con- d 2m sider the model illustrated in Figure 1, where the ball is Equation (3) gives the motion of the ball during contact with represented by its mass m, viscous damping c, and linear the ground and applies only when x · 0. Bounce behav- stiffness k. When the ball is not in contact with the ground, ior, involving deformation, restitution, and then rebound, the equation of motion, assuming no aerodynamic drag, can requires an underdamped solution for which !d > 0 or be written simply as (4km ¡ c2) > 0. mxÄ = ¡mg ; (1) The “steady” or rest solution, applying after the bounces have died out, can be obtained by setting t ! 1 in eq. (3). where x is measured vertically up to the ball’s center of The equilibrium position is mass with x = 0 corresponding to initial contact, i.e., when mg x¤ = ¡ ; (5) the ball just contacts the ground with no deformation. The k and when jxj · jx¤j there will be no further bounces. It B. Stiffness and Damping follows that the number of bounces is finite. The ball properties, k and c, can be determined from the contact time, ¢T , and the coefficient of restitution, e, where ¯ ¯ A. Time of Contact ¯ ¯ ¯x_(¢T )¯ e = ¯ ¯ : (10) The time of contact, ¢T , for the first bounce, shown in x_(0) exaggerated view in Figure 2, is the time from when the The denominator of eq. (10) is simply the velocity of the ball reaches x = 0 after being dropped to the time it first ball prior to contact, v0, and the numerator is the rebound or comes back to x = 0. Mathematically, the contact time is post-impact velocity of the ball, v1. The latter can be found the first finite solution of the equation x(¢T ) = 0, i.e., it by differentiating eq. (7) and imposing the assumption is the minimum non-zero solution of mg k ¿ 1 or alternatively differentiating eq. (8) directly to · ¸ give an expression for the velocity, cg ¡ 2kv0 mg sin(!d¢T ) + cos(!d¢T ) £ cv ³ c ´ 2k!d k 0 µ ¶ x_ = exp ¡ t sin !dt c¢T mg 2m!d 2m exp ¡ ¡ = 0 ; (6) ³ c ´ 2m k ¡v exp ¡ t cos ! t ; (11) 0 2m d which in general has multiple solutions. and then substituting t = ¢T with eq. (9) to give the rebound velocity, µ ¶ First c¼ Bounce v1 =x _(¢T ) = v0 exp ¡ : (12) v = 0 h0 2m!d h v = 0 1 i = 1 Thus, from eq. (10), the coefficient of restitution can be written simply as Height v0 v1 µ ¶ c¼ e = exp ¡ : (13) ∆T v0 v1 2m!d Time By manipulating eqs. (4), (9), and (13), the stiffness and viscous damping can be written, respectively, as, m £ ¤ Fig. 2. Height versus time and exaggerated view at first bounce. k = ¼2 + (ln e)2 (14) (¢T )2 Although eq. (6) is difficult to solve analytically, it can be 2m solved numerically. Alternatively, an approximate solution c = ¡ ln e : (15) can be obtained. Start by writing eq. (3) in the rearranged ¢T form, Assuming k, c and e are constant (independent of the ³ ´ velocity v ), ¢T will be constant for each contact since ! v0 c mg 0 d x = ¡ exp ¡ t sin !dt + £ depends only on the system parameters k, c and m. !d 2m k h ³ c ´ ³ c ´ i exp ¡ t cos !dt + sin !dt ¡ 1 : (7) C. Natural Frequency and Damping Ratio 2m 2m! p The undamped natural frequency, !n = k=m, can be mg Assuming k ¿ 1, which is reasonable for a bouncing expressed from eq. (14) as ball such as a ping-pong ball, the second term on the right- p 1 2 2 hand side in (7) can be neglected and x can be approximated !n = [¼ + (ln e) ] (16) as ¢T ³ ´ The damping ratio, ³, v0 c x = ¡ exp ¡ t sin !dt : (8) s !d 2m ! 2 c d p ³ = 1 ¡ 2 = The contact time, ¢T , can be found as the minimum !n 2 km nonzero solution of eq. (8) set equal to zero giving can be found by substituting eqs. (14) and (15) giving ¼ ln e ¢T = ; (9) ³ = ¡p (17) !d ¼2 + (ln e)2 where !d is specified by eq. (4). Equation (9) represents an Eq. (17) indicates that the damping ratio depends solely on approximate solution for the contact time at the first bounce. the coefficient of restitution. D. Coefficient of Restitution and Time of Contact assuming the contact times at the bounces are identical. For a given ball, the mass m is readily available whereas Noting that the flight time for the ith bounce can be written as T = eT for i ¸ 2 and T = eT , the total flight time the parameters k and c or, alternatively, !n and ³ are i i¡1 1 0 generally unknown. From ¢T and e, which are also un- for the number of bounces n can be calculated using eq. known (but can be found experimentally), k and c can be (22): determined from eqs. (14) and (15), or ! and ³ can be 1 n T = T + T + ¢ ¢ ¢ + T determined from eqs. (16) and (17). flight 2 0 1 n The total number of bounces of the ball, n, and the total 1 n¡1 = T0 + T1(1 + e + ¢ ¢ ¢ + e ) time, Ttotal, that elapses from when the ball is dropped until 2 µ ¶ it comes to rest are two parameters that can be determined 1 1 ¡ en¡1 = T + T e readily in an experiment. They are indicated in the bounce 2 0 0 1 ¡ e µ ¶ history diagram of Figure 3. In the following, it is shown 1 1 + e ¡ 2en = T : that with n and Ttotal assumed known, ¢T and e, and thus 2 0 1 ¡ e k and c, can be determined under the assumption of constant q ¢T and e for all bounces and neglecting aerodynamic drag. 2h0 Since T0 = 2 g , the total flight time for the number of bounces n can be expressed as, Height First Second Third n-th s Bounce Bounce Bounce Bounce µ n ¶ 2h0 1 + e ¡ 2e i = 1 i = 2 i = 3 i = n Tflight = : (24) v = 0 g 1 ¡ e h0 h v = 0 1 Substituting eqs. (20), (23), and (24) into (21) gives v = 0 s h2 h h0 3 T = £ total g Fall Rise Fall Rise Fall Rise · µ ¶ ¸ p n p v v v v v v 1 + e ¡ 2e n 0 1 1 2 2 3 2 + ne ¼2 + (ln e)2 : (25) 1 ¡ e ½T0 T1 T2 ∆T ∆T Time Eq.(25) can be viewed as a single equation for unknown e T total in terms of Ttotal, n, and h0.