Rigid object problem
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The rolling motion of a rigid object is related to many physical phenomena, starting from daily life, such as the motion of bicycle wheel till the unusual and fascinating rotation of a Gyroscope spinning about its axis of symmetry. In the following problem we deal with such a motion and apply the concepts of angular momentum and conservation laws needed to handle such phenomena. Have a joyful reading !
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Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Rigid object problem
- 1. The Problem Three m masses are placed in the corners of an equilateral triangle having a side length of L. The rods that connect the triangle corners are massless and the triangle is placed in rest on a horizontal frictionless table. A bullet with mass m is moving in a speed V0in the positive X axis direction and hits the upper mass of the triangle, as seen in the drawing. After the collision the bullet is moving in a velocity of V= -0.2V0x. What is the velocity of the center of mass of the triangle after the collision. What is the final angular velocity of the triangle relative to the triangle’s center of mass. Qualitative Analysis ● Three masses are placed in the corners of an equilateral triangle. ● A bullet moving in the positive X direction hits the upper mass and continues moving. Quantitative Analysis ● The mass of each mass and the bullet is m. ● The speed of the bullet before collision is V0(VB). ● The speed of the bullet after collision is -0.2V0(UB). Requirements Analysis ● The velocity of the center of mass of the Triangle after collision (UT). ● The angular speed of the triangle relative to its center of mass after collision (ω). Physics Problem Solving https://youtu.be/DKQf8b-T_2w
- 2. Solution for UT V U m Um B = m B + 3 T − .2V ) UV 0 = ( 0 0 + 3 T (1.2V ) .4VUT = 3 1 0 = 0 0 The velocity of the center of mass of the triangle after collision is .4V 0 0 Physics Problem Solving https://youtu.be/DKQf8b-T_2w
- 3. Solution for ω 1 = Iω LT 2 m rI = Σ i i 2 3 r = L 2cos30 4 LT + LB2 = LB1 5 mU LB2 = r B 6 LB1 = mV r 0 4 r(V ) LT = LB1 − LB2 = m 0 − UB = .2V m = 1 0 L 2cos30 = √3 1.2mLV 0 2 (m( ) ) L I = 3 L 2cos30 2 = m 2 1 L ω√3 1.2mLV 0 = m 2 1 ω = L√3 1.2V 0 The angular speed of the triangle is L√3 1.2V 0 Physics Problem Solving https://youtu.be/DKQf8b-T_2w