Angular acceleration problem

[Punchlinegirl]

A rod of length 57.0 cm and mass 1.90 kg is suspended by two strings which are 41.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

I tried using torque= I* alpha
torque= L x f= (.57)(18.62)=10.6
I got from Newton's 2nd law, (9.8)(1.90)
so, 10.6 = I* alpha
I= (1/12)(mL^2)= (1/12)(1.90 * (.57)^2= .051
so, 10.6 = .051 alpha
alpha = 207.8 rad/s^2
alpha= a/L
207.8 = a / .57
a= 118.4 m/s ^2

This isn't right... can someone please help me?

 

[lightgrav]

First, the place that it is rotating around is at one end of the rod.
How far from this is the rod's center-of-mass?

Second,
Newton's 2nd law is "Sum of Forces = ma"
Maybe you mean Newton's 4th Law "Force by gravity = m g = m GM/r^2"

 

[quicknote]

There are a few mistakes in your calculation. First, the torque should be calculated as the product of the force and the perpendicular distance from the point of rotation to the line of action of the force. In this case, the force is the weight of the rod, which is given by mg, and the perpendicular distance is the distance from the point of rotation (the point where the string is cut) to the center of mass of the rod, which is half of its length (0.285 m). So, the torque should be calculated as 0.285 * 1.9 * 9.8 = 5.6 Nm.

Next, the moment of inertia of a rod about its center of mass is given by 1/12 * m * L^2, where m is the mass and L is the length. So, the moment of inertia in this case should be 1/12 * 1.9 * 0.57^2 = 0.012 kgm^2.

Finally, using the equation torque = moment of inertia * angular acceleration, we can solve for the initial angular acceleration of end B:

5.6 = 0.012 * alpha
alpha = 5.6 / 0.012 = 467 rad/s^2

Since the acceleration of end B is directly proportional to the angular acceleration, we can use the equation a = alpha * L to find the initial linear acceleration:

a = 467 * 0.41 = 191.5 m/s^2

So, the magnitude of the initial acceleration of end B is 191.5 m/s^2.