Find time taken by a pendulum to swing by 90degrees

AI Thread Summary
To determine the time taken for a pendulum to swing from a horizontal to a vertical position, the tangential acceleration of the ball is expressed as gcosθ, with θ being the angle from the horizontal. The relationship between tangential acceleration and angular acceleration is established as Lα = gcosθ, leading to the differential equation d²θ/dt² = (g/L)cosθ. The discussion suggests rewriting this in terms of angular velocity (ω) and integrating to find ω in terms of θ. However, it is noted that solving the resulting integral can be complex, and a closed-form solution for a 90-degree swing may not be readily available. The conversation emphasizes the challenges of deriving a solution for this specific pendulum motion.
vikvaryas
Messages
2
Reaction score
0

Homework Statement


A small ball is attached to a massless rod at one end. The other end is hinged such that the rod can swing freely in the vertical plane.
Find the time taken by this system to rotate from horizontal position to vertical position.
Length of rod = L
Acceleration due to gravity = g
mass of ball = m
All surfaces are frictionless.

Homework Equations




The Attempt at a Solution


Tangential acceleration of ball is gcosθ where θ is angle the rod makes with horizontal
Tangential acceleration is equal to Lα where α is rate of change of angular speed
So i have Lα=gcosθ
I am unable to proceed further. what do i do?
 
Physics news on Phys.org
Firstly write the expression for angular acceleration and then try to find angular velocity by integration.
 
Sunil Simha said:
Firstly write the expression for angular acceleration and then try to find angular velocity by integration.

I have obtained a differential equation d2θ/dt2 = (g/L)cosθ
How do i solve this?
 
vikvaryas said:
I have obtained a differential equation d2θ/dt2 = (g/L)cosθ
How do i solve this?

Write it as dω/dt. But dω/dt can also be written as (dω/dθ)*(dθ/dt) or ω*dω/dθ. So now integrate and you'll obtain ω in terms of θ.

Once you get that, ω can be written as dθ/dt and if you integrate the expression relating ω and θ, you can relate θ and t.
 
Sunil Simha said:
Write it as dω/dt. But dω/dt can also be written as (dω/dθ)*(dθ/dt) or ω*dω/dθ. So now integrate and you'll obtain ω in terms of θ.

Once you get that, ω can be written as dθ/dt and if you integrate the expression relating ω and θ, you can relate θ and t.
Easily said, but you end up with a very nasty integral. See http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period . One might hope that there is a closed form solution for the specific case of a 90 degree amplitude, but I'm not aware of such.
 
  • Like
Likes PeroK
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top