Moment of Inertia with varying distance from Centre of Mass

[Ang09]

Homework Statement

A T-shaped card is made of two pieces of thin cardboard sticked together with tape. Each of this cardboard is 0.28 m by 0.04 m. There are several holes along the vertical portion of the T-shape. The T-shaped card is pivoted one of the holes along the vertical portion of the T-shape.

In term of h (where h is the distance between the pivot and the centre of mass of the T-shaped card), what is the moment of inertia of the T-shaped card ?

Relevant Equations

Moment of inertia of a thin card I = 1/12 m(l^2 + b^2) where l and b are length and breadth of card
I = Icom + md^2

h = d1 + 0.08
d1 = h - 0.08

d2 = h + 0.08

I of the vertical portion
= 1/12 m (l^2 + b^2) + md1^2
= 1/12 m (0.28^2 + 0.04^2) + m(h - 0.08)^2

I of the horizontal portion
= 1/12 m (l^2 + b^2) + md2^2
= 1/12 m (0.28^2 + 0.04^2) + m(h + 0.08)^2

The moment of inertia for the whole T-shape about the pivot (of distance h away from centre of mass) =
= 1/12 m (0.28^2 + 0.04^2) + m(h - 0.08)^2 + 1/12 m (0.28^2 + 0.04^2) + m(h + 0.08)^2
= 1/6 m (0.28^2 + 0.04^2) + 2m(h^2 + 0.08^2)
= 2mh^2 + 0.02613 m

Wondering if my calculation is correct ?

 

[haruspex]

How are you defining d1 and d2? Where does the 0.08 come from?

 

[Ang09]

d1 is the distance between pivot and centre of mass of the vertical portion
d2 is the distance between pivot and centre of mass of the horizontal portion
(Refer to picture)

The 0.08 m is derived based on the diagram to relate the centre of mass of the vertical and horizontal portion to the pivot.

 

[Ang09]

The computation of the centre of mass is based on something that i search online.
[Link deleted by the Mentors]

My computation was:-
COM of the horizontal part y1: 0.02 m from the base of the horizontal part
COM of the vertical part y2: (0.14 + 0.04=) 0.18 m from the base of the horizontal part

COM of the T-shaped pendulum
= (Area1 * y1 + Area2 * y2) / (Area1 + Area2)
= (0.28*0.04 *0.02 + 0.28*0.04 *0.018) / (0.28*0.04 + 0.28*0.04)
= 0.10 m
Hence it is 0.08 m away from both of the COM of the horizontal and vertical part.

I guess it may be wrong. May I ask what is the center of mass of the two strips?

By the way, is the computation steps for the moment of inertia of T-shaped pendulum correct? i.e. if i can find the correct center of mass, would the steps above allow me to find the moment of inertia of the T-shaped pendulum in terms of h?

Thank you for your advice in advance.

 

[haruspex]

The computation of the centre of mass is based on something that i search online.
[Link deleted by the Mentors]

View attachment 284264
My computation was:-
COM of the horizontal part y1: 0.02 m from the base of the horizontal part
COM of the vertical part y2: (0.14 + 0.04=) 0.18 m from the base of the horizontal part

COM of the T-shaped pendulum
= (Area1 * y1 + Area2 * y2) / (Area1 + Area2)
= (0.28*0.04 *0.02 + 0.28*0.04 *0.018) / (0.28*0.04 + 0.28*0.04)
= 0.10 m
Hence it is 0.08 m away from both of the COM of the horizontal and vertical part.

I guess it may be wrong. May I ask what is the center of mass of the two strips?

By the way, is the computation steps for the moment of inertia of T-shaped pendulum correct? i.e. if i can find the correct center of mass, would the steps above allow me to find the moment of inertia of the T-shaped pendulum in terms of h?

Thank you for your advice in advance.

I am very sorry - I was writing rubbish.
Let me try again.. give me a few minutes.

Yes, your calculation looks ok, but the question as posted does not define m.
Should it be the mass of the whole shape or the mass of each strip?

 

[Ang09]

Sorry.. i am not clear in my working..

• the m in the 0.02 m and 0.18 m are the unit (metre).
• the m in the moment of inertia is the mass of each strip.

 

[haruspex]

[*]the m in the moment of inertia is the mass of each strip.

Yes, but in post #1 we are asked to find an MoI without being given a mass, neither as "m" nor as a numerical quantity. Does the original question say the mass of each strip is m, that the mass of the whole shape is m, or are you left to make something up?

 

[Ang09]

The original ques did not state the mass of the t-shape pendulum nor the mass of each cardboard. Only inform that the two cardboard are identical/similar. So I assume that each cardboard's mass is m.
Need that when I relate the MoI of the pendulum to its period.

 

[haruspex]

The original ques did not state the mass of the t-shape pendulum nor the mass of each cardboard. Only inform that the two cardboard are identical/similar. So I assume that each cardboard's mass is m.
Need that when I relate the MoI of the pendulum to its period.

Ok.
For finding the period, it should not matter whether you take m as the mass of each or as the combined mass, provided you are consistent.