Derive velocity function for pendulum

[quantum13]

Homework Statement

A regular pendulum is swinging back and forth. Assuming it starts from a horizontal position, find an expression for its tangential velocity.

(I don't know if the question is posed right, I just asked myself it as a challenge)

2. The attempt at a solution
using a freebody diagram, I know that the tangential vector is the tangential component of weight. let Θ be the angle such that the tangential component of weight is mgsinΘ.

F = ma
mgsinΘ = ma
gsinΘ = dv/dt

I know I am supposed to do
v = ∫gsinΘ dt
but I have no clue how to do that except using rotation equations,
Θ = α/2 t^2 + ωt = a/2r t^2 + v/r t = gcosΘ/2r t^2 + v/r t

which leaves me with messy algebra that brings me in circles...

can anyone help me set up this problem?

 

[AEM]

Homework Statement

A regular pendulum is swinging back and forth. Assuming it starts from a horizontal position, find an expression for its tangential velocity.

(I don't know if the question is posed right, I just asked myself it as a challenge)

2. The attempt at a solution
using a freebody diagram, I know that the tangential vector is the tangential component of weight. let Θ be the angle such that the tangential component of weight is mgsinΘ.

F = ma
mgsinΘ = ma
gsinΘ = dv/dt

I know I am supposed to do
v = ∫gsinΘ dt
but I have no clue how to do that except using rotation equations,
Θ = α/2 t^2 + ωt = a/2r t^2 + v/r t = gcosΘ/2r t^2 + v/r t

which leaves me with messy algebra that brings me in circles...

can anyone help me set up this problem?

What you might want to do is write v in terms of theta. If l is the length of the pendulum, then $v = l \frac{d\theta}{dt}$. Then the next thing you should do is restrict you pendulum to small oscillations (small values of theta) that way you can use the standard $sin(\theta) \approx \theta$ approximation. Note two things: you will end up with the second derivative of theta, and the equation for simple harmonic motion. You can use the solution of that equation to get the velocity you want from the relationship between the linear and angular velocity I wrote earlier.

 

[kerimek]

To derive the velocity function for a pendulum, we can start by considering the forces acting on the pendulum bob. The only force acting on the bob is its weight, mg, which can be resolved into two components: a tangential component, mg sinΘ, and a radial component, mg cosΘ.

Using Newton's second law, we can write the equation of motion for the pendulum as:

mg sinΘ = ma

where m is the mass of the bob, a is its acceleration, and Θ is the angle between the tangential component of weight and the vertical axis.

We can also use the relationship between angular and linear acceleration, a = rα, where r is the length of the pendulum and α is the angular acceleration. Substituting this into the equation above, we get:

mgr sinΘ = mrα

Simplifying, we get:

g sinΘ = rα

We also know that the tangential velocity, v, is related to the angular velocity, ω, by the equation v = rω. Therefore, we can rewrite the equation above as:

g sinΘ = r(dω/dt)

Integrating both sides with respect to time, we get:

∫ g sinΘ dt = ∫ r(dω/dt) dt

Integrating the left side gives us:

gt cosΘ + C1

where C1 is a constant of integration. On the right side, we can use the chain rule to simplify the integral:

∫ r(dω/dt) dt = ∫ rω dω = rω^2/2 + C2

where C2 is another constant of integration.

Combining these results, we get:

gt cosΘ + C1 = rω^2/2 + C2

To find the values of C1 and C2, we can use the initial conditions of the problem. Since the pendulum starts from a horizontal position, we know that at t=0, Θ=0 and ω=0. Substituting these values into the equation above, we get:

C1 = C2 = 0

Therefore, the final expression for the tangential velocity of the pendulum is:

v = √(2gr(1-cosΘ))

or, in terms of the length of the pendulum, L:

v = √(