Answer:
a) The resulting angular speed of platform is 1.38 rev/sec
b) The change in kinetic energy of the system is 53 J.
Explanation:
This question is incomplete. The complete question will be:
You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 7.0 k
g
.m
2
a) What is the resulting angular speed of the platform? Answer in units of r
e
v
/
s
.
b)What is the change in kinetic energy of the system? Answer in units of J.
<h3>
ANSWER:</h3>
a)
we know that:
Angular Momentum = L = Iω
From conservation of momentum:
Lo = Lf
(Io) (ωo) = (If) (ωf)
ωf = (Io) (ωo)/(If)
ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)
<u>ωf = 1.38 rev/sec =</u>
b)
ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec
ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec
The kinetic energy for rotational motion is given as:
K.E = (1/2)Iω²
Thus, the change in kinetic energy will be:
ΔK.E = (K.E)f - (K.E)o
ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²
ΔK.E = (1/2)(Ifωf² - Ioωo²)
ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²
<u>ΔK.E = 53 J</u>
