Respuesta :
Using the concepts of centre of mass,we got that 1.525m should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1. 30 m from the left-hand end of the rod.
Let the clamp be at x distance from the left hand end of rod.
Taking left hand as origin , the formula for centre of mass is in x co-ordinates is given by the summation of product of masses and their displacement .
We are given that mass of rod is 1.80Kg and mass of clamp is 2.40kg.
So, applying the formula
x ( cm ) =(m₁ .x ₁+m₂ .x₂)/(m₁+m₂)
We are given that Centre of gravity of the composite object to be 1.30 m
So, we need to find x₁.
After putting the values,
1.30=[(2.4×x₁)+(1.8×1)] / [(2.4+1.8)]
=>1.30×(2.4+1.8)=(2.4×x₁+1.8)
=>1.30×4.2=2.4×x₁+1.8
=>2.4×x₁=1.30×4.2-1.8
=>2.4x₁=5.46-1.8
=> x₁=3.66/2.4
=>x₁=1.525m
Hence, the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1. 30 m from the left-hand end of the rod should be far about 1.525m
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(Complete question) is:
A uniform rod is 2.00 m long and has a mass of 1.80 kg. A 2.40-kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.30 m from the left-hand end of the rod?
