Torque and angular acceleration problem

1. Jul 25, 2005

feather

Hi, I have a problem that I just can't seem to figure out. Here it is:

Four masses are arrnaged as shown below. They are connected by rigid, massless rods of lengths 0.75 m and 0.50 m. What torque must be applied to cause an angular acceleration of 0.75 rad/s^2?

A--------B
l
l
------------(Axis of rotation)
l
l
D--------C

(B and C are connected by a rod, I couldn't get the picture to look right)

A= 4 kg, B=3 kg, C=5 kg, D=2 kg

I know that torque=I(rotational inertia) times angular acceleration, but I don't know what equation for I to use. None of the equations given in the book for I, (hoop, cylinder, sphere, rod, and plate) seem to fit. I'm stuck!

I would appreciate any help!

2. Jul 25, 2005

qbert

the formula you want to look into is
$$I = \sum m_i r_i^2$$

where m_i is the mass of the i'th particle
and r_i is the distance from that particle
to the axis of rotation.

3. Jul 26, 2005

Dr.Brain

Let T be the point of meeting of the axis of rotation and road AD . Then what you need to do is to calculate the M.I around the T point. First calculate M.I of AD around T (how to calculate rod's moment of inertia around the centre?) , then calculate M.I of AB and CD rodad around the axis of rotation using parallel axis theorem. Once you add all those M.I , you get the net M.I which is to be used in T=I.(angular acceleration)

BJ

4. Jul 26, 2005

quasi426

Use the formula given by qbert, it is the equation used to solve the moment of inetia for all the other geometeries listed in your book, such as a sphere, rod and hoop. But when you have just a few point masses their is no need to use the integral and it can be numerically solved by summing the contribution of each mass relative to the axis of rotation. The only reason you can take this approach is because the rod is massless (or can be neglected.)