Problem 2: Experiment 09 Physical Pendulum. Part One: Ruler Pendulum

Transcription

1 Problem : Experiment 9 Physical Pendulum Part One: Ruler Pendulum The ruler has a mass m r =.159 k, a width a =.8 m, a lenth b = 1. m, and the distance from the pivot point to the center of mass is l =.479 m. Enter your measured period into the T meas column of the table below and calculate the other entries usin the formulas and with = 9.85 ms -. T = π l theory / T ideal = π l / I θ 1+ ml 16 θ T meas T theory T ideal.1.5 Part Two: Added Mass Consider the effect of a brass weiht clipped to the ruler. The weiht is shaped like a washer with an outer radius r o =.16 m and an inner radius r i =. m; it has a mass m w =.5 k. It is clipped to the ruler so that the inner hole is over the.5 m mark on the ruler, or l =.479 m from the pivot point. The clip has a mass m c =.86 k and you may assume its center of mass is also over the.5 m mark on the ruler. If you treat the washer and clip as point masses, 3

2 then, as was discussed in the notes for Experiment 8, the combined unit (ruler, weiht and clip) has a moment of inertia about the pivot point I P = m ( a + b ) + m 1 r r c w l + (m + m )d where d = l for this situation. The restorin torque that tries to return the pendulum to a vertical position will be τ = (m l + m d + m ) d sin θ (m l + m d + m ) d θ r c w r c w 1. Use these two expressions to derive an equation of motion for the pendulum and calculate its period T in the small amplitude (sinθ θ ) approximation. Express your answer alebraically in terms of the variables a, b, d, l, m r, m w, m c, and.. Evaluate your result numerically and compare with the value you measured in your experiment. 3. If you treat the brass object washer as a point mass, its moment of inertia about the pivot point P is I w,p = m w l. If the brass object is a washer with an inner radius r i and outer radius r, then moment of inertia about its center of mass iven by I = 1 m r + r o w ( w o i ). If the washer is a solid disc with radius r, the moment of inertia about its center of mass iven by I = 1 w mr w.when this is taken into account, what is the new (and more accurate) expression for I w,p? How many percent does this differ from the simpler expression I w,p = m w l? 4

3 Problem : Experiment 9 Physical Pendulum Part One: Ruler Pendulum The ruler has a mass m r =.159 k, a width a =.8 m, a lenth b = 1. m, and the distance from the pivot point to the center of mass is l =.479 m. Enter your measured period into the T meas column of the table below and calculate the other entries usin the formulas and T = π l theory / T ideal = π l / I θ 1+ ml 16 with = 9.85 ms -. Solution: The first part of the analysis of the experiment is to calculate the moment of inertia about an axis passin throuh the center of mass, perpendicular to the plane formed by the sides of the ruler. In particular, choose Cartesian coordinates with the oriin at the center of mass, and the x-axis alon the lenth, and y-axis alon the width. The mass per unit area σ = mass m 1 =. Area ab The mass element is rotatin about the z-axis in a circular orbit with radius r = ( x + y ) 1, so the moment of inertia about the center of mass of the ruler is 3

4 x = b y = a I cm = (r ) dm = σ,cm ( x + y ) dy body x= b y= a dx We first do the interal in the y-direction, 3 I = m x= b 1 x y + y x = b 3 y a dx = m = 1 a cm y= a x a + dx ab x= b 3 ab x= b 1 We now do the interal in the x-direction = = a + x = m b 1 a + b = m 1 ( b + a x= b ) m I 1 x a x b a S ab 3 1 ab = (.159k ) ((1. m) + (.8m) ) = k m 1 Now use the parallel axis theorem to calculate the moment of inertia about the pivot point, = m 1 ( b + a ) + ml 1 1 cm Usin the data for the ruler, the moment about the pivot point is = k m + (.159 k )(.479 m ) = k m. θ T meas T theory T ideal Part Two: Added Mass Consider the effect of a brass weiht clipped to the ruler. The weiht is shaped like a washer with an outer radius r o =.16 m and an inner radius r i =. m; it has a mass m w =.5 k. It is clipped to the ruler so that the inner hole is over the.5 m mark on the ruler, or l =.479 m from the pivot point. The clip has a mass m c =.86 k and you may assume its center of mass is also over the.5 m mark on the ruler. If you treat the washer and clip as point masses, 4

5 then, as was discussed in the notes for Experiment 8, the combined unit (ruler, weiht and clip) has a moment of inertia about the pivot point I P = m ( a + b ) + m l 1 r + (m + m ) d r c w where d = l for this situation. The restorin torque that tries to return the pendulum to a vertical position will be τ = (m l + md + m d) sin θ (m l + md + m d) θ r c w r c w 1. Use these two expressions to derive an equation of motion for the pendulum and calculate its period T in the small amplitude (sinθ θ ) approximation. Express your answer alebraically in terms of the variables a, b, d, l, m r, m w, m c, and. Answer:. Evaluate your result numerically and compare with the value you measured in your experiment. Solution The moment of inertia of the washer and binder clip treated as point masses about the pivot point is cw = ( m + m,, c w ) l cm ( =.586 k.479 m ) = k m 5

6 Thus the total moment of inertia is total = m (b + a ) + ml + (m c + m ) l 1 cm w cm total = k m k m = k m. So the new period is π T = π ω p cm total 1 k m l m total = π 6.3 (.479 m )(.18 k )(9.85 m s - ) = 1.56 s So the approximation ives ood areement with the data. Displacement [ m ] Measured Period [ s ] Calculated Period T [ s ] If you treat the brass object washer as a point mass, its moment of inertia about the pivot point P is I w,p = m w l. If the brass object is a washer with an inner radius r i and outer radius r, then moment of inertia about its center of mass iven by I = 1 m r + r o w ( w o i ). If the washer is a solid disc with radius r, the moment of inertia about its center of mass iven by I = 1 w mr w. When this is taken into account, what is the new (and more accurate) expression for I w,p? How many percent does this differ from the simpler expression I w,p = m w l? Solution: In Problem 1, you showed that the moment of inertia of a washer about the center of mass is iven by the result, I = 1 m r + r ) cm w ( i, so the total moment of inertia is now total = m (b + a ) + ml + ( m + m ) l + 1 m w (r + r ) cm c w cm i 1 6

7 The moment of inertia of the washer about the center of mass is I cm, w 1 = m r + r ) i = 1 (.5k)((.16m) + (.m) ) = k m. w ( 6 This is neliible compared to the overall moment of inertia. The ratio of the moment of the washer about its center of mass compared to the moment of the washer treated as a point mass about the pivot point is I cm, w k m = wcm k m ml =