Angular acceleration and velocity problem

[ippo90]

Homework Statement

a homogeneous pole AB has a mass m and a length l. the pole makes a 45 deg angle with the horizontal axis, and the pole is being held in resting position by a rope. (see the drawing =P)
find the angular acceleration and angular velocity, when the rope is cut and the pole doesn't slide.

Homework Equations

Eq. for torque: t= Ia or t = r x F
Eq. for the spin: L=Iw or L = r x p

The Attempt at a Solution

tried to find the ang. acceleration by setting the eq. for torque equal each other.

Ia=rxF

I=1/12mr^2, where r is l/2 because the center of mass is there(?).

and

r x F is rmgsin(agle)

sonving this I got (24g/l)sin(a)
which is wrong how should I approach this problem?
hope someone will answer!

 

[anathema]

Hello there,

To find the angular acceleration and angular velocity of the pole, we need to consider the forces and torques acting on it. In this case, we have the weight of the pole acting downwards and the tension force from the rope acting upwards. Since the pole is not sliding, the net force on it must be zero.

Now, let's consider the torque equation. The torque is equal to the moment of inertia (I) times the angular acceleration (α). We can also express it as the cross product of the radius vector (r) and the force (F). In this case, the force is the weight of the pole (mg) and the radius vector is the distance from the pivot point (A) to the center of mass of the pole (l/2). Therefore, our torque equation becomes:

t = Iα = r x F = (l/2) x mg = (1/2)lmg

Since the pole is not sliding, the net force on it must be zero. This means that the weight of the pole must be balanced by the tension force from the rope. Therefore, we can write the following equation:

mg = T

Now, we can substitute this into our torque equation to get:

t = (1/2)lmg = Iα

Solving for α, we get:

α = (1/2)lg/I = (1/2)lg/(1/12)ml^2 = 6g/l

This is the angular acceleration of the pole when the rope is cut.

To find the angular velocity, we can use the equation L = Iω, where L is the angular momentum and ω is the angular velocity. The angular momentum can be expressed as the cross product of the radius vector and the linear momentum (p). In this case, the linear momentum is zero since the pole is not sliding. Therefore, our equation becomes:

L = Iω = r x p = (l/2) x 0 = 0

This means that the angular velocity is also zero when the rope is cut.

I hope this helps! Let me know if you have any further questions.