Moment of inertia and the two theorems on axes · Physics Class 11–12 | StemPy Lab
Course 6.2·Lesson 2 of 3
Moment of inertia and the two theorems on axes
The moment of inertia I=∑miri2
Lesson 2 of 3Show all lessonsHide
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Turning inertia
Mass measures how hard a body is to speed up in a straight line. For turning, what matters is not just the mass, but how far it lies from the axis. That is the moment of inertia:
I=∑miri2
in kg m2. Mass far from the axis counts much more than mass near it.
Five shapes with their axes of rotation. A thin ring, I equals M R squared. A disc, I equals one half M R squared. A thin rod with the axis through its middle at a right angle to it, I equals M L squared over 12. A thin rod with the axis through one end, I equals M L squared over 3. A solid sphere with an axis through its centre, I equals two fifths M R squared.
The radius of gyrationk is the distance at which all the mass would give the same I: I=Mk2.
Two theorems that move the axis
Parallel axes. If Icm is about an axis through the centre of mass, then about any parallel axis a distance d away,
I=Icm+Md2
A disc with a green dashed axis through its centre of mass, and a red dashed axis parallel to it at a distance d. The moment of inertia about the red axis is the moment of inertia about the centre of mass plus M d squared.
Perpendicular axes (for flat bodies only). With x and y in the plane and z at right angles to it, all through one point:
Iz=Ix+Iy
So a disc, with Iz=21MR2, has 41MR2 about any diameter.
You should find the moment of inertia smallest about the middle, and four times as big about an end.
Watch out. The parallel axis theorem must start from the axis through the centre of mass. Moving between two other axes needs two steps.
A worked example
Two 2.0kg balls are joined by a light rod 0.60m long.
About the middle. Each ball is 0.30m away: I=2×2.0kg×(0.30m)2=0.36kg m2.
About one ball. One ball is on the axis, the other 0.60m away: I=2.0kg×(0.60m)2=0.72kg m2.
Check by the theorem.I=Icm+Md2=0.36kgm2+4.0kg×(0.30m)2=0.72kg m2.
Think first. For point masses, add mr2. For standard shapes, start from the table and move the axis with a theorem.
Practise with new numbers
Try it in Python
The program cuts a rod into a thousand small pieces, adds up mr2, and compares it with the formula.
Remember these
I=∑miri2, in kg m2; I=Mk2.
Ring MR2; disc 21MR2; rod 121ML2 (middle), 31ML2 (end); solid sphere 52MR2.
I=Icm+Md2; for flat bodies Iz=Ix+Iy.
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
A 0.5kg ball whirls on a light string 0.8m long. What is its moment of inertia about the centre?
A ring and a disc have the same mass and radius. Which is harder to set spinning about its axis? Why?
What is the moment of inertia of a 4.0kg uniform rod 3.0m long about one end?
A flat square plate has I=0.8kg m2 about an axis through its centre at right angles to it. What is I about an axis in its plane, through its centre, parallel to a side?
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