A Mass-Spring-Damper Model of a Bouncing

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 and the 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 textbooks, not make a connection between the mechanical “primitives” leading to the classical results of vertical projectile . 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 mx¨ + 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 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 ω t + cos ω t × 2kω 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, ωd, is II.MASS-SPRING-DAMPER MODEL 1 p ω = 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 , can requires an underdamped solution for which ωd > 0 or be written simply as (4km − c2) > 0. mx¨ = −mg , (1) The “steady” or rest solution, applying after the bounces have died out, can be obtained by setting t → ∞ in eq. (3). where x is measured vertically up to the ball’s center of The equilibrium 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 |x| ≤ |x∗| 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(ωd∆T ) + cos(ωd∆T ) × cv ³ c ´ 2kωd k 0 µ ¶ x˙ = exp − t sin ωdt c∆T mg 2mωd 2m exp − − = 0 , (6) ³ c ´ 2m k −v exp − t cos ω 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 cπ Bounce v1 =x ˙(∆T ) = v0 exp − . (12) v = 0 h0 2mωd h v = 0 1 i = 1 Thus, from eq. (10), the coefficient of restitution can be written simply as Height v0 v1 µ ¶ cπ e = exp − . (13) ∆T v0 v1 2mωd 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 = π2 + (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 ω v0 c mg 0 d x = − exp − t sin ωdt + × depends only on the system parameters k, c and m. ωd 2m k h ³ c ´ ³ c ´ i exp − t cos ωdt + sin ωdt − 1 . (7) C. Natural Frequency and Damping Ratio 2m 2mω p The undamped natural frequency, ωn = 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 ωn = [π + (ln e) ] (16) as ∆T ³ ´ The damping ratio, ζ, v0 c x = − exp − t sin ωdt . (8) s ωd 2m ω 2 c d √ ζ = 1 − 2 = The contact time, ∆T , can be found as the minimum ωn 2 km nonzero solution of eq. (8) set equal to zero giving can be found by substituting eqs. (14) and (15) giving π ln e ∆T = , (9) ζ = −p (17) ωd π2 + (ln e)2 where ωd 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 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, ωn and ζ 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 ω and ζ 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 ¶ ¸ √ n p v v v v v v 1 + e − 2e n 0 1 1 2 2 3 2 + ne π2 + (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. The latter three quantities can readily be determined experimentally. Fig. 3. Bounce history showing height versus time. E. Approximations For the i-th bounce, the height the ball can reach is It is possible to develop simplified approximate relation- 2i hi = e h0 (18) ships for the case of |(ln e)/π| ¿ 1, which for a ratio of 0.1 or smaller corresponds to 0.73 < e < 1. This case would where h0 is the height when the ball is dropped since vi = be representative of a ball with significant bounce, such as i √ evi−1 = e v0 and vi = 2ghi. For the ball to come to rest, a ping-pong ball. 2n mg For this case, eq. (14) can be approximated as hn = e h0 ≤ (19) k ³ π ´2 k =∼ m , (26) where the upper limit is given by the equilibrium position ∆T of (5). Substituting eq. (14) into the equality of (19) and which itself is an approximation of eq. (9), rearranging gives an expression for the contact time, ∆T , π in terms of unknown e: ∆T =∼ , (27) s ωn h ∆T = en 0 [π2 + (ln e)2] (20) i.e., the contact time at a single bounce is simply π times g the inverse of the undamped natural frequency. The contact The total time is the sum of the total flight time, T , time can also be approximated, from eq. (20), as flight s and the total contact time, T , contact h ∆T =∼ enπ 0 . (28) Ttotal = Tflight + Tcontact (21) g where From eq. (17), it is also possible to write the damping 1 Xn ratio for the case of higher values of e as T = T + T (22) flight 2 0 i ln e i=1 ζ =∼ − (29) π and providing a simple direct connection between the damping Tcontact = n∆T (23) ratio and the coefficient of restitution. Simplification of eq. (25) gives value is slightly higher than that determined for e at the s µ ¶ first bounce based on pre- and post-impact from ∼ 2h0 1 + e the high- digital video images (i.e., by applying eq. Ttotal = (30) g 1 − e (10)). for larger n and e. Eq. (30) does not depend on n, and can It is also possible to determine the coefficient of restitu- be rearranged to find a simple equation for e in terms of tion from the approximate equation (30). From this equa- tion, for Ttotal = 7.5 s, the coefficient of restitution is Ttotal. e = 0.94. III.NUMERICALAND EXPERIMENTAL STUDIES B. Predicted Contact Time A ping-pong ball (Harvard, one-star) was dropped from rest from a measured initial height of 30.5 cm onto a The contact time at a single bounce can be found from (butcher-block top) laboratory bench. The acoustic signals eq. (20) or the approximation of eq. (28). These relations accompanying the ball-table impacts were recorded using a are shown graphically in Figure 5, from which the contact microphone attached to the card of a PC. The method can be determined by inspection given the total number of follows the procedure described in [6]. bounces, n, and the coefficient of restitution, e. For n = 70 From the temporal history of the bounce of and e = 0.93, the predicted contact time ∆T = 3.4 ms. successive impacts, the total number of bounces was de- This value exceeds the contact time of ∆T = 1.0 ms for termined to be n = 70 and the total bounce time was the first bounce measured by the high-speed digital video determined to be Ttotal = 7.5 s. system. The mass of the ball used in the experiment was measured to be m = 2.50g. (The ball used was an older official ball. 5 The rules of the International Table Federation were 4.5 changed in September 2000 and now mandate a 2.7 g ball.) 0.925 0.930 0.935 e=0.940 4 In addition to the acoustic measurement, a high-speed 0.920 digital video (using a Redlake Imaging MotionScope) was 3.5 0.915 taken. 3 0.910 A. Predicted Coefficient of Restitution 2.5 0.905 The coefficient of restitution can be found from eq. (25) 2 e=0.900 given known initial height h0, total time Ttotal, and number Contact Time (ms) 1.5 of bounces, n. The relationship is shown in Figure 4 for the case of h0 = 30.5 cm and indicates that the total time 1 is not significantly dependent on the number of bounces, 0.5 especially for a large number of bounces. 0 50 55 60 65 70 75 80 Number of Bounces, n 12 Fig. 5. Contact time at a single bounce as a function of number of bounces e=0.950 and coefficient of restitution from eq. (20) and from approximation of eq. 10 (28) for a drop height of 30.5 cm.

8 C. Predicted Stiffness and Damping 0.925 Values of the linear stiffness and the viscous damping 6 coefficient of an equivalent mass-spring-damper model of a 0.900 ball can be determined. Total Time (s) 4 0.875 0.850 An expression for the stiffness is given in eq. (14) and in simplified approximate form in eq. (26). Figure 6 2 graphically depicts these relationships in terms of k/m for e=0.700 the range of coefficient of restitution 0.40 ≤ e ≤ 0.95 for 0 several values of contact time. The approximate equation 20 30 40 50 60 70 80 90 100 Number of Bounces, n (26) provides a highly accurate prediction of the result from Fig. 4. Total time as a function of number of bounces and coefficient of eq. (14), showing only slight deviation at smaller values of restitution from eq. (25) for a drop height of 30.5 cm. e. For the case of the ping-pong ball dropped from an initial Figure 4 provides a means to identify by inspection the height of 30.5 cm and with ∆T determined to be 3.4 ms, 5 2 coefficient of restitution. In particular, for Ttotal = 7.5 s k/m = 8.5 × 10 s and is not a function of e. For m = and n = 70, the coefficient of restitution is e = 0.93. This 2.5g, then the stiffness k = 2.1 N/mm (or kPa). As indicated 8 10 For the case of ∆T = 1.0 ms and e = 0.93, ωn = 3100

∆T=0.5 ms rad/s or 500 Hz. From eq. (17) or the approximation of eq. (29) it is Dashed Line = Approximation possible to predict the damping ratio. For e = 0.93, the ∆T=1.0 ms damping ratio ζ = 0.023. The small value of damping ratio 7 10 indicates a very lightly underdamped system.

) 1.5 ms 2 IV. DISCUSSION 2.0 ms 2.5 ms

k/m (s The total time from the when the ball is dropped until 3.0 ms 6 when it comes to rest is comprised of two phases: flight 10 times and contact times. Although the total contact time ∆T=4.0 ms summed for all bounces is a small fraction of the total flight time (approximately 3 percent), it is included in the model. 3.5 ms The analytical development assumes constant mass- 5 m k c 10 spring-damper model parameters, , , and , and constant 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 coefficient of restitution, e. A consequence of assuming that Coefficient of Restitution, e these parameters are constant is a constant contact time, Fig. 6. Stiffness divided by mass as a function of coefficient of restitution and contact time from eq. (14) and for approximation from eq. (26). ∆T , at each bounce. The analysis neglects aerodynamic effects, which occur in reality. By not accounting for aerodynamic drag of the above, ∆T was measured to be 1.0 ms from the high-speed ball during flight, the approach gives a higher coefficient of digital video. With this value, k/m = 1.0 × 107 s2 and the restitution than otherwise would be predicted. stiffness k = 25 N/mm (or kPa). The approach predicts a contact time three times greater An equation for the damping coefficient c was developed than that found by an independent method (3.4 ms vs 1.0 in eq. (15), and is plotted in Figure 7 as c/m as a function ms using high-speed digital video). Reconciling this large of both e and ∆T , showing clear dependence on both. difference requires further study into the errors resulting For e = 0.93 and ∆T = 3.4 ms, c/m = 43 s−1 and for from the underlying assumptions, namely, neglecting aero- m = 2.5g then the damping coefficient c = 0.11 N·s/m. For dynamic drag and adopting a linear, fixed mass-spring- the case of e = 0.93 and ∆T = 1.0 ms, c/m = 145 s−1 damper model. and c = 0.36 N·s/m. It is noted that the equivalent damping Several observations can be made: (i) the larger the is predicated on knowledge of e and the value of the contact contact time ∆T , the smaller the stiffness k and the larger time ∆T . the damping c, (ii) the larger the coefficient of restitution e, the smaller the damping c, (iii) the coefficient of restitution 1000 e does not strongly influence the stiffness k, (iv) the larger 900 the coefficient of restitution e, the larger the total time, and ∆T=4.0 ms (v) the number of bounces n (assuming n > 20) does not 800 strongly influence the total time. 700 V. CLOSING 600 ∆ T=3.5 ms This paper examines relationships bridging linear equiv- 500 alent model parameters, namely the mass, stiffness, and c/m (1/s) 400 ∆T=3.0 ms damping of a bouncing ball, with the classical concept of 2.5 ms 300 coefficient of restitution and time of contact between a ball 2.0 ms 1.5 ms and a surface. Under the assumption of no aerodynamic 200 drag and constant coefficient of restitution for all bounces, ∆T=0.5 ms 100 the derivation shows that the stiffness and damping, or al-

0 ternatively the natural frequency and damping ratio, can be 0.7 0.75 0.8 0.85 0.9 0.95 Coefficient of Restitution, e expressed explicitly in terms of the coefficient of restitution and time of contact. The formulation also considers the Fig. 7. Damping coefficient divided by mass as a function of coefficient of restitution and contact time from eq. (15). special case for bouncing involving higher values of the coefficient of restitution for which simple approximate expressions can be derived for parameters of the ball model. D. Predicted Natural Frequency and Damping Ratio The results of an experimental test are used to provide From eq. (16) or the approximation from a rearrangement predictions of the equivalent stiffness and damping, as well of eq. (27) it is possible to find the natural frequency. For as natural frequency and damping ratio, and coefficient of ∆T = 3.4 ms and e = 0.93, ωn = 920 rad/s or 150 Hz. restitution for a bouncing ping-pong ball. REFERENCES [1] Flansburg, L. and Hudnut, K., 1979, “Dynamic Solutions for Linear Elastic Collisions,” American Journal of Physics, 47, pp. 911-914. [2] Pauchard, L. and Rica, S., 1998, “Contact and Compression of Elastic Spherical Shells: the Physics of a Ping-Pong Ball,” Philosophical Magazine B, 78(2), pp. 225-233. [3] Cross, C., 1999, “The Bounce of a Ball,” American Journal of Physics, 67(3), pp. 222-227. [4] Nagurka, M.L., 2003, “Aerodynamic Effects in a Dropped Ping-Pong Ball Experiment,” International Journal of Engineering Education, 19(4), pp. 623-630. [5] Wu, C., Li, L. and Thornton, C., 2003, “Rebound Behavior of Spheres for Plastic Impacts,” International Journal of Impact Engineering, 28, pp. 929-946. [6] Nagurka, M.L., 2002, “A Simple Dynamics Experiment Based on Acoustic Emission,” Mechatronics, 12(2), pp. 229-239.