How to find the Kinetic energy at position 2 in terms of the angular speed? The equation I know 1 is: KE=mv² +1w². However, other solutions include using the instantaneous center method. Why is the mv² term not there anymore when IC method is used? Does this problem both have translation and rotation in the first place? The answer should be 0.1667w² A spring of constant k = 520 N/m is attached to end A of a 4.00-kg rod AB, which is constrained to move along slots cut in a vertical plane as shown. When 0 = 0, rod A is at rest and the spring is undeformed (Position 1). Consider the instant when 0 = 45.0° as Position 2. Hint: I_rod = 1/12 mL^2 A L = 0.500 m 0 B

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How to find the Kinetic energy at position 2 in terms of the angular speed? The equation I know
is: KE=mv² +1w². However, other solutions include using the instantaneous center method.
2
1
Why is the mv² term not there anymore when IC method is used? Does this problem both
have translation and rotation in the first place? The answer should be 0.1667w²
A spring of constant k = 520 N/m is attached to end A of a 4.00-kg rod AB, which
is constrained to move along slots cut in a vertical plane as shown. When 0 = 0,
rod A is at rest and the spring is undeformed (Position 1). Consider the instant
when 0 = 45.0° as Position 2. Hint: I_rod = 1/12 mL^2
A
L = 0.500 m
0
B
Transcribed Image Text:How to find the Kinetic energy at position 2 in terms of the angular speed? The equation I know is: KE=mv² +1w². However, other solutions include using the instantaneous center method. 2 1 Why is the mv² term not there anymore when IC method is used? Does this problem both have translation and rotation in the first place? The answer should be 0.1667w² A spring of constant k = 520 N/m is attached to end A of a 4.00-kg rod AB, which is constrained to move along slots cut in a vertical plane as shown. When 0 = 0, rod A is at rest and the spring is undeformed (Position 1). Consider the instant when 0 = 45.0° as Position 2. Hint: I_rod = 1/12 mL^2 A L = 0.500 m 0 B
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