Classical Mechanics-Moments of Inertia and Torques

[sclatters]

Homework Statement

a)Two people are holding the ends of a plank of length l and mass M. Show that, if one suddenly let's go, the initial acceleration of the free end (a_D) 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).

Homework Equations

I*angular acceleration=torque=rxF

parallel axis theorem for the moment of inertia

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.

 

[Doc Al]

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.

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

 

[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?

 

[Doc Al]

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?

You got it.

 

[HalfLostAndSearching]

As a scientist, the first thing to do when faced with a problem like this is to carefully analyze the given information and identify any relevant equations or concepts that can be applied. In this case, the key concepts are classical mechanics, moments of inertia, and torques. The given information includes the length and mass of the plank, as well as the initial acceleration of the free end and the load supported by the other person.

To solve part a), we can use the equation for the moment of inertia of a uniform rod about its centre of mass, I=1/12(Ml2). This equation tells us the distribution of mass along the length of the plank, which is important for calculating the moment of inertia. We can also use the equation torque=rxF, where r is the distance from the pivot point to the point where the force is applied, and F is the force itself. In this case, the force is the weight of the plank, which is equal to Mg.

To solve part b), we can use the parallel axis theorem for the moment of inertia, which states that the moment of inertia of an object about any axis parallel to its centre of mass is equal to the moment of inertia about the centre of mass plus the product of the mass of the object and the square of the distance between the two axes. This will allow us to calculate the moment of inertia of the plank about the pivot point, which is necessary for calculating the torque and the change in load on the other person's hand.

In order to solve both parts of the question, we also need to make some assumptions, such as assuming that the plank is a rigid body and that the forces acting on it are balanced. We also need to use the concept of conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque.

Using these concepts and equations, we can solve both parts of the question and provide a complete and accurate response. It is important to carefully analyze the given information and make appropriate assumptions in order to come to a correct solution.