The pendulum is released from rest with θ = 30deg

[Alexanddros81]

Homework Statement

13.54 The pentulum is released from rest with θ = 30deg. (a) Derive the equation of motion
using θ as the independent variable. (b) Determine the speed of the bob as a function of θ.

The solutions given in the textbok are a) $\ddot θ = -4.905sinθ rad/s^2$
b) $6.26\sqrt {cosθ - 0.866} m/s$

Homework Equations

The Attempt at a Solution

[/B]
I understand something is not correct.
I guess the positive direction for the bob now is to the left.

 

[Orodruin]

Where did the 7.5 come from? You have not been given the tension, you have been given that the bob is released from rest.

 

[rude man]

part (a) think torque = I times dω/dt
part (b) think energy conservation.

 

[Alexanddros81]

Hi! After doing some research on the web I came up with the following solution (by an engineer)
I don't understand though why he took $a_T$ to be $-rα$. Does it has to do with the
the tangential and radial axes sense?
He mentions $F=ma_T$ as an inertial force.

 

[haruspex]

took $a_T$ to be −rα

If you start with arc length = radius x arc angle then differentiate twice wrt time you will get this. The sign depends on your conventions. In vectors, $\vec {a_T}=\vec r\times\ddot{\vec\theta}$.
You can get the second result (velocity) by integrating or by work conservation.

 

[rude man]

Hi! After doing some research on the web I came up with the following solution (by an engineer)
I don't understand though why he took $a_T$ to be $-rα$. Does it has to do with the
the tangential and radial axes sense?
He mentions $F=ma_T$ as an inertial force.

View attachment 212931

View attachment 212937

Your last result v = 6.26√(cosθ - cos30deg) is correct.