The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration ke=0.4 m. The spring's unstretched length is Lo=1.0 m. The stiffness coefficient of the spring is k-2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is 8-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 0=0°. The spring's length at the state 2 is L2=4 m. (4) The elastic potential energy at the potion 1 is_ HULKU 2₂ State 2 G m State 1 (N-m) (two decimal places)

Elements Of Electromagnetics
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The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration kG=0.4 m. The spring’s unstretched length is L0=1.0 m. The stiffness coefficient of the spring is k=2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is θ=30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is θ=0°. The spring’s length at the state 2 is L2=4 m.

 

(4) The elastic potential energy at the potion 1 is_______(N·m) (two decimal places)

 

The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the
wheel is 0.6m. The radius of gyration ke=0.4 m. The spring's unstretched length is Lo=1.0 m.
The stiffness coefficient of the spring is k-2.0 N/m. The wheel is released from rest at the
state 1 when the angle between the spring and the vertical direction is 8-30°. The wheel rolls
without slipping and passes the position at the state 2 when the angle is 0=0°. The spring's
length at the state 2 is L2=4 m.
(4) The elastic potential energy at the potion 1 is_
HULKU
2₂
State 2
G
m
State 1
(N-m) (two decimal places)
Transcribed Image Text:The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration ke=0.4 m. The spring's unstretched length is Lo=1.0 m. The stiffness coefficient of the spring is k-2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is 8-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 0=0°. The spring's length at the state 2 is L2=4 m. (4) The elastic potential energy at the potion 1 is_ HULKU 2₂ State 2 G m State 1 (N-m) (two decimal places)
Expert Solution
Step 1

Given Data:

Mass of Wheel m=20 kg

Radius of Wheel R=0.6 m

Radius of gyration KG=0.4 m

Unstretched length of spring L0=1 m

Stiffness coefficient of spring k=2 Nm

The angle between the spring and vertical direction θ=30°

Length of spring at state 2 L2=4 m

 

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