Respuesta :
Answer:
a) 1.54 rad/s
b) 2212 J
c) 1030J
Explanation:
Given that
Radius of turntable, r = 2.5 m
Total mass of turntable, M = 130 kg
Speed of rotation of turntable, w = 3.3 rad/s
Mass of parachutist, m = 74 kg
To solve this, first, we make an assumption:
That the turntable and the parachutist are a system. Then we use law of conservation of momentum i.e the angular momentum is the same before and after landing. Thus,
I1w1 = I2w2
I1 = moment of inertia of the disk
I1 = 1/2MR²
I1 = 1/2 * 130 * 2.5²
I1 = 65 * 6.25
I1 = 406.25 kgm²
I2 = moment of inertia of the parachute and the turntable
I2 = 1/2MR² + mR²
I2 = 406.25 + 74 * 2.5²
I2 = 406.25 + 74 * 6.25
I2 = 406.25 + 462.5
I2 = 868.75 kgm²
I1w1 = I2w2
w2 = I1w1 / I2
w2 = (406.25 * 3.3) / 868.75
w2 = 1340.625 / 868.75
w2 = 1.54 rad/s
k1 = 1/2.I1.w1²
k1 = 1/2 * 406.25 * 3.3²
k1 = 1/2 * 406.25 * 10.89
k1 = 2212 J
k2 = 1/2.I2.w2²
k2 = 1/2 * 868.75 * 1.54²
k2 = 1/2 * 868.75 * 2.3716
k2 = 1030 J
The KE is different because of the negative work done by the parachutist on the turntable, during the softlanding
Answer:
a) 1.54 rad/s
b) 2212 J
c) 1030J
Explanation:
Given:
Radius 'r' = 2.5 m
Total mass of turnable 'M'= 130 kg
Speed of rotation of turntable ' w' = 3.3 rad/s
Mass of parachutist 'm' = 74 kg
A) Assuming the system of turntable and the parachutist. As we know that the system of turn table and parachutist is isolated and there is no external force on this system.
By applying law of conservation of momentum i.e the angular momentum is the same before and after landing. Therefore,
[tex]I_{1}[/tex][tex]w_{1}[/tex]= [tex]I_{2[/tex][tex]w_{2[/tex]
Where,
[tex]I_{1}[/tex] is moment of inertia of the disk and [tex]I_{2[/tex] is moment of inertia of the parachute and the turntable
[tex]I_{1}[/tex]= 1/2MR² => 1/2 x 130 x 2.5²
[tex]I_{1}[/tex]= 65 x 6.25
[tex]I_{1}[/tex]= 406.25 kgm²
[tex]I_{2[/tex] = 1/2MR² + mR² => 406.25 + (74 x 2.5²)
[tex]I_{2[/tex] = 406.25 + 74 x 6.25 =>406.25 + 462.5
[tex]I_{2[/tex] = 868.75 kgm²
[tex]I_{1}[/tex][tex]w_{1}[/tex]= [tex]I_{2[/tex][tex]w_{2[/tex]
[tex]w_{2[/tex] = [tex]I_{1}[/tex][tex]w_{1}[/tex]/[tex]I_{2[/tex]
[tex]w_{2[/tex] = (406.25 x 3.3) / 868.75 =>1340.625 / 868.75
[tex]w_{2[/tex] = 1.54 rad/s
B) For initial kinetic energy :
[tex]k_{1[/tex] = 1/2[tex]I_{1[/tex][tex]w_{1[/tex]² => 1/2 x 406.25 x 3.3²
[tex]k_{1[/tex] = 1/2 x 406.25 x 10.89
[tex]k_{1[/tex] = 2212 J
C)For final kinetic energy:
[tex]k_{2[/tex] = 1/2[tex]I_{2[/tex] [tex]w_{2[/tex]² => 1/2 x 868.75 x 1.54²
[tex]k_{2[/tex] = 1/2 x 868.75 x 2.3716
[tex]k_{2[/tex] = 1030 J
D) Because of the negative work done by the parachutist on the turntable, during the softlanding we have different KE. Energy is decreased due to friction between parachutist and the disc.