Kinetic energy = (1/2) (mass) (speed)²
= (1/2) (1.4 kg) (22.5 m/s)²
= (0.7 kg) (506.25 m²/s² )
= 354.375 kg-m²/s² = 354.375 joules .
This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.
I was about to say: because people generally get comfortable with
what they think they know, and don't like the discomfort of being told
that they have to change something they're comfortable with.
But then I thought about it a little bit more, and I have a different answer.
"Society" might initially reject a new scientific theory, because 'society'
is totally unequipped to render judgement of any kind regarding any
development in Science.
First of all, 'Society' is a thing that's made of a bunch of people, so it's
inherently unequipped to deal with scientific news. Anything that 'Society'
decides has a lot of the mob psychology in it, and a public opinion poll or
a popularity contest are terrible ways to evaluate a scientific discovery.
Second, let's face it. The main ingredient that comprises 'Society' ... people ...
are generally uneducated, unknowledgeable, unqualified, and clueless in the
substance, the history, and the methods of scientific inquiry and reporting.
There may be very good reasons that some particular a new scientific theory
should be rejected, or at least seriously questioned. But believe me, 'Society'
doesn't have them.
That's pretty much why.
Answer:
ω = 12.023 rad/s
α = 222.61 rad/s²
Explanation:
We are given;
ω0 = 2.37 rad/s, t = 0 sec
ω =?, t = 0.22 sec
α =?
θ = 57°
From formulas,
Tangential acceleration; a_t = rα
Normal acceleration; a_n = rω²
tan θ = a_t/a_n
Thus; tan θ = rα/rω² = α/ω²
tan θ = α/ω²
α = ω²tan θ
Now, α = dω/dt
So; dω/dt = ω²tan θ
Rearranging, we have;
dω/ω² = dt × tan θ
Integrating both sides, we have;
(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ
This gives;
-1[(1/ω_o) - (1/ω)] = t(tan θ)
Thus;
ω = ω_o/(1 - (ω_o × t × tan θ))
While;
α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²
Thus, plugging in the relevant values;
ω = 2.37/(1 - (2.37 × 0.22 × tan 57))
ω = 12.023 rad/s
Also;
α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²
α = 8.64926751525/0.03885408979 = 222.61 rad/s²
