someone spinns with an angular velocity of 4.55rad/s with arms held out. his rotational inertia is 2.7kgm^2. whn the he pulls his arms in his angular velocity increases to 7.60rad/s. what is his rotary ek?
Use conservation of angular momentum to find his final moment of inertia, I2.
I1*w1 = I2*w2
I2 = I1*(4.55/7.60) = 1.616 kg m^2
The final rotational KE is
(1/2) I2*(w2)^2 = (1/2)*1.616*(7.60)^2
(joules)
thanks
i got 2.790 is that correct
To calculate the rotational kinetic energy (Rotary Ek) of the person, we can use the formula:
Rotary Ek = (1/2) * I * ω^2
where:
Rotary Ek is the rotational kinetic energy
I is the rotational inertia
ω is the angular velocity
Given:
Angular velocity (initial) = 4.55 rad/s
Rotational inertia = 2.7 kgm^2
Angular velocity (final) = 7.60 rad/s
First, let's calculate the initial rotational kinetic energy:
Rotary Ek (initial) = (1/2) * I * ω(initial)^2
= (1/2) * 2.7 kgm^2 * (4.55 rad/s)^2
= 0.5 * 2.7 kgm^2 * 20.7025 rad^2/s^2
≈ 55.8167 J (Joules)
Next, let's calculate the final rotational kinetic energy:
Rotary Ek (final) = (1/2) * I * ω(final)^2
= (1/2) * 2.7 kgm^2 * (7.60 rad/s)^2
= 0.5 * 2.7 kgm^2 * 57.76 rad^2/s^2
≈ 77.904 J (Joules)
Therefore, the person's initial rotational kinetic energy is approximately 55.8167 J and their final rotational kinetic energy is approximately 77.904 J.