Question

Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an Axis perpendicular to its plane and passing through a point on it will be​

Answer 1

Answer:

It is said that I is the moment of inertia of the disc about one of its diameter. ... However, the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is equal to $\dfrac{M{{R}^{2}}}{2}$, where M and R are the mass and radius of the disc respectively.

Answer 2

Given :-

Momentum of inertia = I = 1.2 kg -m^2

Rotetional kinetic energy, Kt = 1500J

Angular acceleration

α = 25 rad/s^2, ωo = 0, t =?

Solution :-

Kinetic energy of rotation is given by

Krot = 1/2 Iω^2

∴ ω = √2Kr /I = √ 2×1500/1.2

= 50 rad/s

Now, from equation of rotational motion

ω = ωo + αt

t = (ω - ωo)/ α

t = 50- 0)/ 25 = 2s

∴ Therefore the answer is 2s!

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