g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
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Ответ:
339.3 N
Explanation:
First, we start by converting the units.
1 rev/s = 2π rad/s, so
0.6 rev/s = 2π * 0.6 rad/s
0.6 rev/s = 1.2π rad/s
0.6 rev/s = 3.77 rad/s
Now we apply the equation of motion,
W(f) = w(o) + αt
3.77 = 0 + α * 2
3.77 = 2α
α = 3.77/2
α = 1.885 rad/s²
Torque = I * α
Torque = F * r
This means that
I * α = F * r, where I = 1/2mr²
Substituting for I, we have
1/2mr²α = F * r, making F the subject of formula, we have
F = 1/2mrα, then we substitute for the values
F = 1/2 * 240 * 1.5 * 1.885
F = 678.6 / 2
F = 339.3 N
Ответ:
It is a type of map used to map out the geographical change over time. I believe
Explanation: