A uniform rod of mass 2 kg and length 1 m is placed on a frictionless horizontal surface. Two forces, F 1 = 10 N and F 2 = 6 N , are applied at its ends in opposite directions along the length of

Question

A uniform rod of mass 2 kg and length 1 m is placed on a frictionless horizontal surface. Two forces, F 1 ​ = 10 N and F 2 ​ = 6 N , are applied at its ends in opposite directions along the length of the rod. What is the magnitude of the acceleration of the center of mass of the rod in m / s 2 ?

TestHub · Accepted Answer

The acceleration of the center of mass of a system is determined by the net external force acting on the system and its total mass. The two forces $F_{1} = 10 N$ and $F_{2} = 6 N$ are applied in opposite directions. Therefore, the net external force $F_{n e t}$ acting on the rod is the difference between these forces. $F_{n e t} = ∣ F_{1} - F_{2} ∣ = ∣10 N - 6 N ∣ = 4 N$

The mass of the rod is $M = 2 k g$ . The acceleration of the center of mass $a_{COM}$ is given by Newton's second law: $F_{n e t} = M a_{COM}$ . $a_{COM} = \frac{F _{n e t}}{M} = \frac{4 N}{2 k g} = 2 m / s^{2}$

Physics - Center of Mass Question with Solution | TestHub

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