Question

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

Answer 1

the result
`a_(cm)` = g .The correct answer is d

Step-by-step explanation:

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

`x_(cm)` = 1 / M ∑
`x_(i) m_(i)`

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