Question

Calculate the moment of inertia of thin uniform rod of mass 100 g and length 50 cm about an axis perpendicular to its length and passing through 1.)its centre 2.)one end .

Answer 1

Answer:

1) 0.002

2) 0.008

Explanation:

100 g = 0.1 kg

50 cm = 0.5 m

formula :

1. its centre : ml^2 / 12 = 0.1*(0.5)^2 / 12 = 0.00208333333
2. its end : ml^2 / 3 = 0.1 *(0.5)^2 / 3 = 0.00833333333

Answer 2

Answer:

moment of inertia of rod about its centre

Ic= 3 x 10 ^-3 kg - m²

moment of inertia of rod about its one end

I= 12 x 10 ^-3  kg - m²  

Explanation:

Moment of inertia, in physics, quantitative degree of the rotational inertia of a frame—i.e., the competition that the frame famous to having its velocity of rotation approximately an axis altered via way of means of the utility of a torque (turning force). The axis can be inner or outside and can or won't be fixed.

What is Perpendicular Axis Theorem? M.O.I of a 2-dimensional item approximately an axis passing perpendicularly from it's far same to the sum of the M.O.I of the item approximately 2 jointly perpendicular axes mendacity withinside the aircraft of the item.

so the given problem

second of inertia of rod approximately its centre  Ic= ml²/12

for m = 100 g  and l= 60 cm

Ic= 100 x 10^-3 x 60 x 60 x 10^-4 / 12

Ic= 3 x 10 ^-3 kg - m²

moment of inertia of rod about its one end

I=ml²/ 3

I=100 x 10^-3 x 60 x 60 x 10^-4 / 3

I= 12 x 10 ^-3  kg - m²

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