Having a little trouble with this physics problem.. could you explain how you get the answer? please..?

Question

A rod with mass M = 1.4 kg and length L = 1.2 m is mounted on a central pivot . A metal ball of mass m = 0.7 kg is attached to one end of the rod. You may treat the metal ball as a point mass. The system is oriented in the vertical plane and gravity is acting. The rod initially makes an angle θ = 27 with respect to the horizontal. The rod is released from rest. What is the angular acceleration of the rod immediately after it is released?

Answer 1

Well, both the rod and the ball will experience the pull of gravity, directed downwards, at thier respective centers of gravity. however, because of the support the pivot gives, they can only accelerate in a direction perpendicular to the rod. You need to first determine the component of the gravitational force for each element in this body that is directed along this vector.

Then, I think you need to determine the torque developed (F*d) and compare with the moment of inertia of this system about the pivot. From there, you should be able to determine the angular acceleration.

I hope that helps!

Answer 2

You need to look at the torques acting on:
1. the center of mass of the rod above the pivot
2. the center of mass of the rod below the pivot
3. the mass of the ball.

Half the mass of the rod is abvoe the pivot point. The center of this mass is locate a distance L/2/2 = L/4 above the pivot

The torque due to gravity is Tup = - M/2*g*L/4*sin(q) where q = angle between the rod and the vertical (gravity). I picked the "-" sign assuming this torque is in teh counterclockwise direction (top of the rod is to the left).

The torque on the bottom half is Tbot = +(M/2 *L/4 + m*L/2)*g*sin(a) where m = mass of ball and a = angle between rod and vertical. Now you can easily show a+q = 180 so sin (q) = sin(180-a) = sin(a).

The toques sum:
I*alpha = Tup + Tbot = - M/2*g*L/4*sin(a)+(M/2 *L/4 + m*L/2)*g*sin(a) = m*L/2*g*sin(a)

Now I = m*(L/2)^2

So

alpha =2*g*sin(a)/L

Finally a = 90 -27 = 63deg

alpha = 14.55 rad/s^2

Answer 3

sure. i became in church with the college (i became 10 years previous), sitting a bench in between the rows, with my fellow college scholars sat alongside the bench. Then a instructor of mine who became very fat unexpectedly sat down swiftly on the top of the bench, making the entire bench strikes (it became like shockwaves have been sent down the bench, and the bench wasn't small the two, it ought to extra wholesome a minimum of 9 people and her). all the college scholars around me have been scared stiff of her yet I unexpectedly began uncontrollably guffawing from the midsection of the bench. collectively as all people around me became gazing me. the instructor then remarked that "it wasn't that humorous" in a stern and pass way and then proceeded to "supply me detention for performing in an irrelevant way". yet my goodness it became hilarious (even in spite of the undeniable fact that persons witnessing the form purely began to chuckle as quickly as that they had have been given out of the church).