Pendulum problem concerning the spring constant

In summary, the conversation discusses the velocity and acceleration of a mass on the end of a rigid, massless arm at various angles. The problem involves finding the tension in the pendulum using equations of motion in the form of a spring.
  • #1
Benzoate
422
0

Homework Statement



The velocity of the mass at the end of the arm is v^2=2*g*cos(theta) , where theta is the angle the arm makes with the vertical, and we assumed that the arm is rigid and massless. ... I will post the link that has the remaining part of my question since its difficult for me to right "r hat" and "theta hat" Just using character symbols.

http://courses.ncsu.edu/py411/lec/001/

Remaining part of the question should be on second page of my homework.




Homework Equations



v^2=2*G*L*cos(theta)

The Attempt at a Solution



differentiating v^2 , I get v=-g*sin(theta)

a=(2g *cos(theta))"r-hat" - (g sin(theta))"theta hat", a being the acceleration. My acceleration will be in the "r-hat" and "theta hat" polar coordinates rather than cartesian coordinates. since theta is soooooooo small, cos(theta) approximates to 1 and sin(theta) approximates to 0. Therefore , a = 2g"r hat" +0"theta hat"=> a=2g"rhat". I don't understand why the problem just say that a= K*x rather than defined K*x to be the second derivative of the position vector, since I do not need to derived the position vector twice to get the acceleration vector. Mot sure what K real symbolizes is this problem. In this problem, they want me to show the equation of motion in the form of a spring , so I guess then k=sqrt(g/L), but would I need to write out K explicitly? In addition, the Forces of the pendulum would be in the theta hat and r hat direction .

r hat: T-mg cos(theta)=m*a

theta hat:=-mg*sin(theta)=m*a

So I should be able to find my Tension now right? I'm not even sure if the Forces of the pendulum are relevant to this problem.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Benzoate said:

Homework Statement



The velocity of the mass at the end of the arm is v^2=2*g*cos(theta) , where theta is the angle the arm makes with the vertical, and we assumed that the arm is rigid and massless. ... I will post the link that has the remaining part of my question since its difficult for me to right "r hat" and "theta hat" Just using character symbols.

http://courses.ncsu.edu/py411/lec/001/

Remaining part of the question should be on second page of my homework.




Homework Equations



v^2=2*G*L*cos(theta)

The Attempt at a Solution



differentiating v^2 , I get v=-g*sin(theta)

a=(2g *cos(theta))"r-hat" - (g sin(theta))"theta hat", a being the acceleration. My acceleration will be in the "r-hat" and "theta hat" polar coordinates rather than cartesian coordinates. since theta is soooooooo small, cos(theta) approximates to 1 and sin(theta) approximates to 0. Therefore , a = 2g"r hat" +0"theta hat"=> a=2g"rhat". I don't understand why the problem just say that a= K*x rather than defined K*x to be the second derivative of the position vector, since I do not need to derived the position vector twice to get the acceleration vector. Mot sure what K real symbolizes is this problem. In this problem, they want me to show the equation of motion in the form of a spring , so I guess then k=sqrt(g/L), but would I need to write out K explicitly? In addition, the Forces of the pendulum would be in the theta hat and r hat direction .

r hat: T-mg cos(theta)=m*a

theta hat:=-mg*sin(theta)=m*a

So I should be able to find my Tension now right? I'm not even sure if the Forces of the pendulum are relevant to this problem.

How come no one has answered my question yet? Is it too long? Is there something wrong with my wording?
 
Last edited by a moderator:
  • #3
Benzoate said:
How come no one has answered my question yet? Is it too long? Is there something wrong with my wording?

Well first its only been 13 hours since you posted and secondly, you never linked directly to your homework questions so people are going to be confused when they don't find it. I'll link to it here: http://courses.ncsu.edu/py411/lec/001/PY411-HW2-08.pdf

The pendulum is subject to an acceleration and its position is given in terms of theta by [itex] L \theta[/itex]. The acceleration of the bob along that path must be equal to the acceleration you derived previously (in your notes). Note that you can ignore the r components.
 
Last edited by a moderator:

FAQ: Pendulum problem concerning the spring constant

What is a pendulum problem concerning the spring constant?

A pendulum problem concerning the spring constant is a physics problem that involves a pendulum attached to a spring. The spring constant is a measure of the stiffness of the spring and affects the period of the pendulum's motion.

How does the spring constant affect the period of a pendulum?

The spring constant affects the period of a pendulum by determining how much force the spring exerts on the pendulum. A higher spring constant means a stiffer spring and a shorter period, while a lower spring constant results in a longer period.

What is the relationship between the spring constant and the frequency of a pendulum?

The spring constant and the frequency of a pendulum have an inverse relationship. As the spring constant increases, the frequency decreases, and vice versa.

How can the spring constant be calculated in a pendulum problem?

The spring constant can be calculated in a pendulum problem by using Hooke's Law, which states that the force exerted by a spring is directly proportional to its spring constant and the displacement of the spring from its equilibrium position.

How does the length of the spring impact the pendulum's motion in a pendulum problem?

The length of the spring does not directly impact the pendulum's motion in a pendulum problem. However, the length of the spring can indirectly affect the period of the pendulum by changing the equilibrium position of the pendulum's swing.

Back
Top