The formula we use
here is:
radial acceleration =
ω^2 * R <span>
110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>
<span>and since: ω = 2pi*f --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
<span>= 36,714.77 rpm </span></span>
I believe it is speed.
Hope this helps!
Once the car is on top of the hill it contains potential energy witch means it is storing enough energy to slide the hill until acted on. Once the car moves and slides down the hill it creates kinetic energy witch means it's in motion. The car then turns the kinetic energy into mechanical energy witch means it's working. I'm not sure if that helped but good luck!
Answer:
ωf = 4.53 rad/s
Explanation:
By conservation of the angular momentum:
Ib*ωb = (Ib + Ic)*ωf
Where
Ib is the inertia of the ball
ωb is the initial angular velocity of the ball
Ic is the inertia of the catcher
ωf is the final angular velocity of the system
We need to calculate first Ib, Ic, ωb:
ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s
Now, ωf will be:
