# Classical Mechanics-Moments of Inertia and Torques

1. Dec 3, 2013

### sclatters

1. The problem statement, all variables and given/known data
a)Two people are holding the ends of a plank of length l and mass M. Show that, if one suddenly lets go, the initial acceleration of the free end (aD) is 3g/2. (7 marks).

Moment of inertia, I, of the plank about its centre of mass is given by I=1/12(Ml2)

b)Show also that the load supported by the other person falls from Mg/2 to Mg/4. (3 Marks).
2. Relevant equations
I*angular acceleration=torque=rxF

parallel axis theorem for the moment of inertia
3. The attempt at a solution
I have completed the first 7 marks with no difficulty but am really struggling with how to set part b) up and what assumptions I need to make. I've spent hours and hours on this and have ended up going round in circles.

I think I may need to calculate the linear acceleration of the centre of the plank and from this deduce the new load on the other hand. However I have not made much progress with this part of the question.

2. Dec 3, 2013

### Staff: Mentor

That's exactly what you have to do. What forces act on the body? Apply Newton's 2nd law.

3. Dec 3, 2013

### sclatters

So I need to calculate the linear acceleration of the centre of the plank then apply newtons 2nd law to this? This would give the net force on the plank? After this I could deduce that the the net force=mg-(the load on the other hand? Would this work?

4. Dec 3, 2013

You got it.