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 thewheel 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 thestate 1 when the angle between the spring and the vertical direction is 8-30°. The wheel rollswithout slipping and passes the position at the state 2 when the angle is 0=0°. The spring'slength at the state 2 is L2=4 m.(4) The elastic potential energy at the potion 1 is_HULKU2₂State 2GmState 1(N-m) (two decimal places)

Answered: The wheel is attached to the spring.…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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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 k_G=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)

Expert Solution

Step 1

Given Data:

Mass of Wheel $m = 20 kg$

Radius of Wheel $R = 0.6 m$

Radius of gyration $K_{G} = 0.4 m$

Unstretched length of spring $L_{0} = 1 m$

Stiffness coefficient of spring $k = 2 \frac{N}{m}$

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

Length of spring at state 2 $L_{2} = 4 m$

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