Question : A man stands on a rotating platform, with his arms streched horizontally holding a 5kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kgm². What is the new angular speed?
Doubt by Jaskirat
Solution :
Mass of the body (m) = 5 kg
Initial Angular Velocity (ω1) = 2π(30/60) rad/s
=π rad/s
=π rad/s
Moment of Inertia of Man + Platform = 7.6 kg/m²
r1=90 cm = 0.9 m
r2=20 cm = 0.2 m
Total initial momentum (I1)
= 7.6+2×5×(0.9)²]
= 7.6+10×(0.81)
= 7.6+2×5×(0.9)²]
= 7.6+10×(0.81)
=7.6+8.1
= 15.7 kgm²
= 15.7 kgm²
Total final momentum (I2)
= 7.6+2×5×(0.2)²
= 7.6+10×0.04
= 7.6+0.4
= 8 kgm²
= 7.6+10×0.04
= 7.6+0.4
= 8 kgm²
Using the Law of conservation of Angular Momentum
Iω=constant
I1ω1=I2ω2
ω2=I1ω1/I2
ω2=(I1/I2)×ω1
ω2=(15.7/8)π
ω2= (15.7×3.14)/8
ω2=49.298/8
ω2=6.162 rad/s
or
ω2=58.875 revolutions per minute.
ω2=I1ω1/I2
ω2=(I1/I2)×ω1
ω2=(15.7/8)π
ω2= (15.7×3.14)/8
ω2=49.298/8
ω2=6.162 rad/s
or
ω2=58.875 revolutions per minute.