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Answer. We know that center of mass will move under the action of force of gravity So $\mathbf{r_c}=\mathbf{v_c}t+\frac{1}{2}\mathbf{g}t^2$ Or $\mathbf{r_c}=\frac{m_1\mathbf{v_1}+m_2\mathbf{v_2}}{m_1+m_2}t+\frac{1}{2}\mathbf{g}t^2$ Multiple Choice Questions Question 23 Two masses m 1 and m 2 separated by thin rod of length L.
Locate the position of center of mass of the two point masses (i) from the origin and (ii) from 3 kg mass. Solution. Let us take, m1 = 3 kg and m2= 5 kg. (i) To find center of mass from the origin: The point masses are at positions, x1 = 4 m, x2 = 8 m from the origin along X axis. The center of mass xCM can be obtained using equation 5.4.
Find the center of mass of a uniform thin hoop (or ring) of mass M and radius r. Strategy First, the hoop’s symmetry suggests the center of mass should be at its geometric center. If we define our coordinate system such that the origin is located at the center of the hoop, the integral should evaluate to zero.
Give an explanation in terms of external forces and center-of-mass concepts. The net external Force on the system is zero, so there is charr in the position the center O-R mass.
Find the distance of the centre of mass of the lamina from AB. Answer all questions. A hot air balloon moves vertically upwards with a constant velocity. When the balloon is at. height of 30 metres above ground level, a box of mass 5 kg is released from the balloon.
Consider the following mass distribution: 5 kg at (0,0), 3 kg at (0,4) and 4 kg at (3,0). All coordinates are measured in meters. Where should a fourth mass of 8 kg be placed so that the center of mass of the arrangement is at (0,0)?
----- Problem 2 ----- Find the COM of a solid uniform right triangle. Answers 1. x_COM = 0.5786 m 2. x_COM = (2/3)*b for the "base" side on the x-axis, and the hypotenuse slanted in 1st quadrant and passing through the origin.
