12 तना State 2 Ita 77 State 1 datum

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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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.

Ignore the spring's mass.

(1) If the datum for gravitational potential energy is set as shown below, the the gravitational potential energy of the wheel at the state 1 is___ N m(two decimal places)

(2) If the datum for gravitional potential energ is set as shown below, the gravitational potential energy of the wheel at the state 2 is___  N m (two decimal places)

(3) At state 1, how long the spring is stretched from its unstretched state (length difference):________(m) (two decimal places)

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

(5)  At state 2, how long the spring is stretched from its unstretched state (length difference):________(m) (two decimal places)

(6) The elastic potential energy of the spring at the state 2 is_______(N·m) (two decimal places) 

(7) The instantaneous center of zero velocity (IC) of the wheel at state 1 is

(8) The mass moment of inertial of the wheel about its mass center G is I=_________(kg·m) (two decimal places)

(9) The mass moment of inertial of the wheel about its IC center at state 1 is IIC =_________(kg·m) (two decimal places)

(10) The total kinetic energy of the system at the state1 is:________ (N·m) (two decimal places)

(11) Apply the theory of work-energy  to the whole system during state 1 and state 2, and calculate the angular velocity of the wheel at the state 2: _______(rad/s) (two decimal places)

12) Based on result from (11), the kinetic energy of the wheel at the state 2 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 kc-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 0-30°. 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.
Ignore the spring's mass.
L2
46411
#
State 2
L₁
-IG-
State 1
datum
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 kc-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 0-30°. 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. Ignore the spring's mass. L2 46411 # State 2 L₁ -IG- State 1 datum
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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.

Ignore the spring's mass.

(5)  At state 2, how long the spring is stretched from its unstretched state (length difference):________(m) (two decimal places)

(6) The elastic potential energy of the spring at the state 2 is_______(N·m) (two decimal places) 

(7) The instantaneous center of zero velocity (IC) of the wheel at state 1 is

(8) The mass moment of inertial of the wheel about its mass center G is I=_________(kg·m) (two decimal places)

(9) The mass moment of inertial of the wheel about its IC center at state 1 is IIC =_________(kg·m) (two decimal places)

(10) The total kinetic energy of the system at the state1 is:________ (N·m) (two decimal places)

(11) Apply the theory of work-energy  to the whole system during state 1 and state 2, and calculate the angular velocity of the wheel at the state 2: _______(rad/s) (two decimal places)

12) Based on result from (11), the kinetic energy of the wheel at the state 2 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 K-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 0-30°. 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.
Ignore the spring's mass.
L2
14411
-1-(--6G
#H
State 2
State 1
datum
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 K-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 0-30°. 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. Ignore the spring's mass. L2 14411 -1-(--6G #H State 2 State 1 datum
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