A rectangle made of two uniform bars lengths L, mass A, and two of length W, mass B, oscillates with SHM about an axis at one corner.

Related BrainMass Solutions

• The Conservation of Angular Momentum and Final Angular Velocity

  A framework consists of two thin, uniform bars of mass M= 12 kg, length L= 3 m, and two thin, uniform bars of mass m= 6 kg, length D= 3 m, forming a rectangle of length L width D.

• Moment of Inertia of a thin uniform rod by integration

  5512 Moment of Inertia of a thin uniform rod by integration A thin, uniform rod has length L and mass M. a. Find by integration its moment of inertia about an axis perpendicular to the rod at one end. b.

• Torques in equilibrium on a leaning ladder.

  The weight W, of the uniform bar acts vertically down at its center of mass, and has a moment arm of length (.5 L cos 45). Step 1.

• Lagrangian of two particles connected by a spring

  Extension of the spring, x oscillates with a frequency of w = sqrt (k/M) where M = m1m2/(m1+m2) Unknown constants A, B, C and D can be found using boundary conditions.

• Practice problems - Rolling, Torque , and Angular Momentum

  (figure below ) A uniform rod (length = 0.60 meters, mass = 1.0 kg ) rotates about and axis through one end with a rotational inertia of 0.12 kg * m^2 .

• Finding the Ratio of Final Period to Initial Period

  6780 Finding the Ratio of Final Period to Initial Period A uniform sphere, mass M= 84 kg, radius R, has two uniform, thin, flexible bars each with mass m= 25 kg, and length half a circumference, wrapped on its surface as shown in Fig. 1.

• Practice problems with rotational inertia - 7 Problems

  Moment of inertia of the body = MI of rod + MI of hoop (both about the axis of rotation) MI rod = (1/3) m L2 MI of a thin rod of mass m and length L rotating about an axis perpendicular to its length and passing through one end.

• Angular Acceleration Alpha of Uniform Board

  For total torque, add the two clockwise torques T1 = (m g )(.5 L) and T2 = (F) (L/3). Note 5. The moment of inertia of a uniform board mass m length L, about an axis at one end is given by: (2) I= (1/3) m L^2 Note 6.

• Simple Harmonic Motion Mass Hanging Spring.

  Using (3) to get w, and Xm as above, you should get: ANSWER x = (.15 m) cos(7 t) The expert examines SHM mass hanging springs. A vertical axis with origins at the equilibrium level and positive locations are found.