How Does Pendulum Speed Change with Angle and Initial Velocity?

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Homework Help Overview

The discussion revolves around a pendulum problem involving the relationship between the angle of release, initial velocity, and the speed of the pendulum bob at various positions. The problem specifically examines the conditions under which the pendulum can swing to a horizontal position.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of energy conservation principles, questioning the relevance of specific speed values at different points in the pendulum's motion. There is discussion about the minimum initial velocity required for the pendulum to reach a horizontal position.

Discussion Status

Participants are actively engaging with the problem, with some expressing confusion over the application of certain values and others clarifying the conditions necessary for the pendulum's motion. There is no explicit consensus, but guidance is being offered regarding the relevance of initial conditions.

Contextual Notes

Some participants note that the initial speed value of 8.35 m/s is not applicable to the second part of the question, indicating a need to reassess the assumptions made in the problem setup.

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[SOLVED] Pendulum Momentum Question

Homework Statement



Figure 8-34 shows a pendulum of length L = 1.25 m. Its bob has speed v0 when the cord makes an angle theta0 = 40 degrees with the vertical.

What is the speed of the bob when it is in its lowest position if v0 = 8 m/s?

The speed is 8.35 m/s.

What is the least value that v0 can have if the pendulum is to swing down and then up to a horizontal position?

Homework Equations



W + KE1 + PE1 = KE2 + PE2

The Attempt at a Solution



W + KE1 + PE1 = KE2 + PE2:

0 + (1/2)m(8.57 m/s)^2 + 0 = (1/2)m v^2 + m(9.8 m/s^2)1.25m

v= 6 or 7 something. The answer in the book is 4.33 m/s
 
Last edited:
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Shackleford said:

Homework Statement



Figure 8-34 shows a pendulum of length L = 1.25 m. Its bob has speed v0 when the cord makes an angle theta0 = 40 degrees with the vertical.

What is the speed of the bob when it is in its lowest position if v0 = 8 m/s?

The speed is 8.35 m/s.

What is the least value that v0 can have if the pendulum is to swing down and then up to a horizontal position?

Homework Equations



W + KE1 + PE1 = KE2 + PE2

The Attempt at a Solution



W + KE1 + PE1 = KE2 + PE2:

0 + (1/2)m(8.57 m/s)^2 + 0 = (1/2)m v^2 + m(9.8 m/s^2)1.25m

v= 6 or 7 something. The answer in the book is 4.33 m/s

Use 8.35 m/s.
 
physixguru said:
Use 8.35 m/s.

Sorry. I forgot to put the 8.35 in there. But, it still doesn't give me the correct answer. I get 6.74 m/s.
 
The final kinetic energy will be zero when the pendulum reaches the horizontal when considering the minimum initial velocity.

The 8.35m/s does not apply here, since that value is based on the first part of the question and is irrelevant to this part. You are trying to find the velocity of the pendulum needed at 40 degrees to the vertical that will just get you to the horizontal position on the other side.
 
hage567 said:
The final kinetic energy will be zero when the pendulum reaches the horizontal when considering the minimum initial velocity.

The 8.35m/s does not apply here, since that value is based on the first part of the question and is irrelevant to this part. You are trying to find the velocity of the pendulum needed at 40 degrees to the vertical that will just get you to the horizontal position on the other side.

Oh, hello! Man. I know I can work it now. I'll do it after I eat. lol.
 
Got it. Thanks. Sometimes, I just read over the problem too quickly and assume I read it correctly.
 

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