A juggler has three objects in the air simultaneously. The masses of the objects are m, 2m, and 3m. At a particular instant, their respective vertical velocities are v, 0, and -v, where positive velocity is upward.1) What is the magnitude of the acceleration of the centre of mass of the three-object system at this instant?a) zerob) g/3c) g/2d) ge) g/6

the result   = g   .The correct answer is d

Explanation:

For this problem let's start by finding the center of mass of objects

          = 1 / M ∑

we apply this equation to our case

        M = m + 2m + 3m

       M = 6m

        x_{cm} = 1 / 6m (x₁ m + x₂ 2m + x₃ 3m)

        x_{cm} = 1/6 (x₁ + 2 x₂ + 3x₃)

We apply Newton's second law to this center of mass

         dp / dt = M a

         d (M v_{cm}) / dt = M a

let's find the speed of the center of mass

        v_{cm} = 1 / 6m (m v + 2m 0 - 3m v)

        v_{cm} = ⅓ v

         M d v_{cm}/ dt = M a

         d v_{cm} / dt = a

at the center of mass the external force of the system is applied, which is the force of gravity

         dv_{cm} / dt = g

therefore the result

               a = g

The correct answer is d

the answer to this question is 60 n

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.