Example 11.6 need help with the last part of the problem 5. 1.32/2 points | Prev

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Example 11.6 need help with the last part of the problem

5. 1.32/2 points | Previous Answers SerPSE9 11 AE.006 Example 11.6 The Seesaw A father of mass mf and his daughter of mass md sit on opposite ends of a seesaw at equal distances from the pivot at the center (see figure). The seesaw is modeled as a rigid rod of mass M and length and is pivoted without friction. At a given moment, the combination rotates in a vertical plane with an angular speed (A) Find an expression for the magnitude of the system's angular momentum (B) Find an expression for the magnitude of the angular acceleration of the system when the seesaw makes an angle with the horizontal A father and daughter demonstrate angular momentum on a seesaw SOLVE IT (A) Find an expression for the magnitude of the system's angular momentum Conceptualize Identify the z axis through O as the axis of rotation in the figure. The rotating system has angular momentum about that axis. Categorize Ignore any movement of arms or legs of the father and daughter and model them both as particles. The system is therefore modeled as a rigid object. This first part of the example is categorized as a substitution problem The moment of inertia of the system equals the sum of the moments of inertia of the three components: the seesaw and the two individuals. We can refer to the table "Moments of Inertia of Homogeneous Rigid Objects with Different Geometries," to obtain the expression for the moment of inertia of the rod and use the particle expression 1 = mr2 for each person. p2 (M Find the total moment of inertia of the system about the z axis through Q 12 2 2

Explanation / Answer

about the center of mass of person,

Icm = M R^2 / 2 = 63 x 0.4^2 / 2 = 5.04 kg m^2

moment of inertia about the center of seesaw, I = Icm + m d^2

I = 5.04 + 63 (3.5- 0.4)^2

I1 = 610.47 kg m^2

for daughter,

I2 = (25 x 0.4^2 / 2) + 25(3.5 - 0.4)^2

I2 = 242.25 kg m^2

Isystem = 610.47 + 242.25 + (10 x 7^2 / 12)

= 893.55 kg m^2

while considering them point mass,

Isystem' = (10 x 7^2 / 12) + (25 + 63)(3.5^2)

= 1118.83 kg m^2

%error = ((1118.83 - 893.55) / 893.55 ) x 100

= 25.2 %

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