Moment if inertia of a point mass

[i_island0]

Homework Statement

I am struggling with a simple point.
Suppose a particle of mass 'm' is located at (a, b) in XY plane (a,b > 0). Gravity is in -Y direction. The particle is to let fall under gravity. Suppose it falls by a height 'h' after some time 't'. I have two questions.
(i) What is moment of inertia of the point mass about the point (0, b)
(ii) What is the moment of inertia of the point mass about the axis passing through O parallel to Z axis.

Homework Equations

The Attempt at a Solution

My answers are:
(i) m (a^2 + h^2)
(ii) ma^2

Comment if possible, thanks in advance.

 

[alphysicist]

Hi i_island0,

Homework Statement

I am struggling with a simple point.
Suppose a particle of mass 'm' is located at (a, b) in XY plane (a,b > 0). Gravity is in -Y direction. The particle is to let fall under gravity. Suppose it falls by a height 'h' after some time 't'. I have two questions.
(i) What is moment of inertia of the point mass about the point (0, b)
(ii) What is the moment of inertia of the point mass about the axis passing through O parallel to Z axis.

Homework Equations

The Attempt at a Solution

My answers are:
(i) m (a^2 + h^2)

That looks right to me.

(ii) ma^2

I don't believe this is correct; how did you arrive at this answer?

When the question speaks about the axis through point O parallel to the Z-axis, do you just mean that the rotation axis is along the Z-axis? (That is, point O is the origin, right?) Notice that your answer (ma^2) would mean that the particle is a distance (a) away from the rotation axis under question.

 

[i_island0]

firstly, the particle is not rotating about point O it is just falling under gravity.
Secondly, let's look at another problem.
Q. Consider an Atwood machine with pulley having mass M, radius R and sufficient friction such that the thread doesn't slip on it. The masses on the two sides are m1 and m2. (Let m1 > m2). Initially the system is at rest. What will happen after time 't'.

Let 'v' be the speed of each block after time 't'.

Let's write their angular momentum: L = m1 v R + m2 v R + Iw.

Rate of change of angular momentum is the net torque.Thus,
Torque = dL/dt = m1 a R + m2 a R + I(alpha) = m1 (alpha.R) R + m2 (alpha.R) R + I(alpha)
= [m1.R^2 + m2.R^2 + I](alpha)

Can't we now say that net torque = (Total MI)(alpha)

Thus, MI of m1 about the center of the pulley = m1R^2

 

[alphysicist]

firstly, the particle is not rotating about point O it is just falling under gravity.

I know the particle is not rotating in a circle about point O, but it's straight line motion is moving it about point O.

The thing is to consider the position vector, pointing from point O to the particle. Since the position vector is changing it's angle around point O, that is the motion about point O. This picture is important for considering your next problem.

Secondly, let's look at another problem.
Q. Consider an Atwood machine with pulley having mass M, radius R and sufficient friction such that the thread doesn't slip on it. The masses on the two sides are m1 and m2. (Let m1 > m2). Initially the system is at rest. What will happen after time 't'.

Let 'v' be the speed of each block after time 't'.

Let's write their angular momentum: L = m1 v R + m2 v R + Iw.

Rate of change of angular momentum is the net torque.Thus,
Torque = dL/dt = m1 a R + m2 a R + I(alpha) = m1 (alpha.R) R + m2 (alpha.R) R + I(alpha)
= [m1.R^2 + m2.R^2 + I](alpha)

Up to this point everything is okay. But when you used:

a --> (alpha) R

remember that you were doing nothing more than relating the linear acceleration of the mass to the angular acceleration of the pulley. The important point is that alpha is not the angular acceleration of the mass.

Can't we now say that net torque = (Total MI)(alpha)

You would need to say something like:

net torque = (total MI) (alpha')

because this angular acceleration alpha' is not the same as the alpha from the previous equation.

(Recall that alpha described how the pulley mass moved around its axis--so setting alpha'=alpha would mean that all parts of the system moved around the axis the same way. But as you pointed out, the masses are not rotating in circles around the axis, so they will have a different angular acceleration, and so the total system angular acceleration will be different.)

Thus, MI of m1 about the center of the pulley = m1R^2

Just as in your previous problem the MI of m1 will be

(m1) (distance from pulley axis)^2

and this will change with time.