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A Physical Pendulum with Damping Due to Sliding Friction

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A Physical Pendulum with Damping Due to Sliding Friction A Physical Pendulum With Damping Due to Sliding Friction A physical pendulum is described where the damping of the motion is due to sliding friction in the support. The motion is shown to consist of damped oscillations, with the amplitude decreasing linearly with time until the motion abruptly stops. The objective of the related experimental study is to show that the observed motion is an good agreement with the predicted equations of motion. 2 I. INTRODUCTION small so that we may assume ¹k << 1. We also assume that the initial angular amplitude of the swing is much We consider a pendulum made up of a length, L, of a less than one radian, that is: jθ0j << 1, where the an- 2"x4" wooden board suspended by a horizontal circular gle is in radians. In this limit we may replace sin θ ' θ rod of radius, R, that passes through a square hole cut and cos θ ' 1. One can write two simultaneous equations through the board. The size of the sides of the square relating the components of force parallel and perpendic- hole are barely larger than the diameter of the rod. As ular to the board to the corresponding components of the the board swings as a pendulum, the points of contact acceleration of the center of mass of the rod. For small between the square hole and the rod must slide around angles of rotation, the acceleration terms are small and the rod. This creates a sliding friction that provides the these two equations can be solved for the force on the main source of damping of the motion. The con¯guration board due to the supporting rod. From these forces one is similar to the situation shown in Figure 1. can derive the torque that will damp the motion of the swinging board as, ¿f ' ¡¹kMgR sgn(dθ=dt): (1) Forces on the Physical Pendulum This torque always tends to slow down the pendulum N t since it always tend to turn the object in the opposite direction to the direction it is moving. Tt To describe the oscillation of the pendulum, we use Ts Ns I® = ¿i, where I is the moment of inertia of the rigid body about the axis of rotation, ® = d2θ=dt2 is the an- P gular acceleration of the rotating rigid body, and the ¿i are the torques due to the external forces acting on the board. One of these forces is due to the force of gravity, CM which gives a torque equal to ¡Mgh sin θ. In the small angle limit this becomes ¡Mghθ, where θ is the angle between the vertical and the swinging board, M is the mass of the board, h is the distance between the axis of θ rotation and the center of mass of the board, and g is the Mg acceleration of a freely falling body under the action of gravity. Using the parallel axis theorem for moments of inertia, we ¯nd L2 I = M h2 + : (2) 12 µ ¶ FIG. 1. The forces acting on the physical pendulum. Solving for ®, we ¯nd 2 In this experiment the sonic range¯nder is used to d θ gh 2 = ¡ 2 L2 θ (3) record the oscillatory motion of the pendulum in an ascii dt Ãh + 12 ! ¯le. The measured motion should then be compared in detail with the motion predicted by Newton's laws. R gh ¡¹ 2 sgn(dθ=dt): k h 2 L µ ¶ Ãh + 12 ! II. APPLICATION OF NEWTON'S LAWS TO Newton's laws applied to the rotation of a rigid body have THE PENDULUM led to a relation between θ and it's derivatives. This re- lation only holds for maximum angles of swing which are We consider the limit where the coe±cient of sliding less than 20o relative to the vertical equilibrium position. friction between the rod and the square hole is relatively To simplify this relation we let, 1 gh !2 = ; (4) III. COMPARING THEORY WITH h2 + L2=12 EXPERIMENT and R ¢θ0 = ¡4¹ : (5) k h µ ¶ Equation (3) becomes It is then sensible to ¯t the motion data to the follow- ing functional form 2 d θ ¢θ0 + !2θ = !2sgn(dθ=dt): (6) dt2 4 µ ¶ y = (a + bx) sin(cx + d) + e; a + bx > 0; (13) We now use the di®erential equation for θ to deter- mine how the pendulum moves. Suppose that at t = 0 the pendulum starts from rest with θ(t = 0) = θ0. As the and, pendulum swings toward the center sgn(dθ=dt) = ¡1, so that until the motion stops at some negative value of θ, we have y = e; a + bx · 0; (14) 2 d θ ¢θ0 + !2θ = ¡ !2: (7) dt2 4 µ ¶ The most general function of time which satis¯es the last where y is the distance of some point on the pendulum equation is from the range¯nder, x is the time in the language of the curve ¯tting program d¯ta.exe, a is the initial am- ¢θ0 θ = A cos(!t + ®) ¡ ; (8) plitude of the motion, b is a negative number that takes 4 the sliding frictional damping into account, c it the same where A and ® are constants to be determined by the as !, d is a phase factor that depends on when the initial conditions: θ(0) = θ0, and (dθ=dt)t=0 = 0. Apply- clock was started, and e is the distance from the ori- ing the second condition we ¯nd that ® = 0. The ¯rst gin of displacements to the equilibrium point. This is condition gives θ0 = A ¡ (¢θ0=4). We have for the ¯rst exactly choice # 16 with d¯ta.exe. We should compare half of a period of oscillation, the value that d¯ta.exe gives for c with what is calculated for ! from Eq. (4). The ¯t parameters determined by ¢θ0 ¢θ0 θ = θ0 + cos(!t) ¡ : (9) d¯ta.exe are contained at the end of the ¯le generated by 4 4 the program. The value of b should be compared with µ ¶ 0 0 When !t = ¼ the object stops at the largest negative 2¹kR!L =(¼h), where L is the distance from the axis of value of θ. We examine how much the amplitude of os- rotation of the point on the pendulum being observed by cillation decreases in a half-period. the range¯nder. To determine this e®ective point, mea- sure the angle amplitude when the pendulum is started 0 0 ¢θ0 ¢θ0 ¢θ0 and determine L so that a = L θ0. To compare b with θ(¼=!) = ¡ θ0 + ¡ = ¡θ0 ¡ : (10) 4 4 2 theory you must make careful measurements of R and µ ¶ the coe±cient of kinetic friction between two similar sur- We see that in one-half of a period that the amplitude faces. Even this is a bit crude, since ¯nding a perfect decreases by ¢θ0=2. This decrease does not depend on match between the two surfaces may be crude. the amplitude. This is the decrease in amplitude that will occur in any half period, ¼=!. Thus, the ampli- You should record the motion for at least three cases. tude decreases at a constant rate with time. This rate is The di®erences between the parameters for the three the change in amplitude ¢θ0=2 divided by ¼=!, the time cases will give a good idea about the error in the mea- during which this change occurs. surements. A major test of the theory lies in how well dθ0(t) ¢θ0! the observed data agrees with Eqs. (13) and (14). This = = ¡2¹ (R=h)(!=¼): (11) dt 2¼ k is the main prediction of Newton's laws. In addition to the ¯t parameters d¯ta.exe writes the root mean square Thus, the pendulum oscillates back and forth with the di®erence between the data and the ¯t to the output ¯le. same period that would describe the motion without fric- Part of this deviation is not a flaw in the theory, but is tion. However, in each swing of the pendulum the am- due to the use of a di®use reflector with the range¯nder plitude decreases by the ¯xed amount, y¢θ0. This can in order not to lose the signal when the angle of reflection be put into a single equation, would be such that the reflection would miss the detec- 2¹kR tor. This would otherwise occur for the larger amplitude θ = θ0 ¡ !t cos(!t + ®): (12) ¼h oscillations. The major subtle points and sources of error µ ¶ should be discussed in your laboratory report. 2 IV. AN EXAMPLE OF DATA ANALYSIS FIG. 2. Measured position as a function of time compared with a curve ¯t with Equations (13) and (14) Finally, we consider a case where the pendulum was This ¯gure shows the kind of data and the agreement pulled back and started from rest with the data being with Eq. (13) that is expected with this study. To com- taken by the Sonic Range¯nder. The recorded motion is pare with the results of the data¯t we use L = 66 cm, shown in Figure 2. h = 24:5 cm, and R = 0:64 cm. We estimate the coe±- cient of friction of smooth steel against smooth steel that is not oiled to be ¹k = 0:3. Calculating from Eq. (3) we get ! = 4:99 /s, and 2¹kR!=(¼h) = 0:025 /s. If we take the distance from the axis of rotation down to the point where the reflection occurs to be 40 cm, we have L0 = 40 cm, so that our estimate of b in the ¯t to the data is 10.00 around b = ¡1:0 cm/s.
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