Question

Fig.3 shows three 10 g particles that have been glued to a rod of length L = 6 cm and negligible mass. The assembly can rotate around a perpendicular axis through point O at the left end. a) How much work is required to change the rotational rate from 40 to 60 rad/s?
b) If the outermost particle is removed, by what percentage does the rotational inertia of the assembly around the rotation axis decrease?

Answer 1

(a) The amount of work required to change the rotational rate is 0.0112 J.

(b) The decrease in the rotational inertia when the outermost particle is removed is 64.29%.

Moment of inertia of the rod

The moment of inertia of the rod from the axis of rotation is calculated as follows;

I = md² + m(2d)² + m(3d)²

where;

• m is mass = 10 g = 0.01 kg
• d = 3 equal division of the length

d = 6/3 = 2 cm = 0.02 m

I = md²(1 + 2² + 3²)

I = 14md²

I = 14(0.01)(0.02)²

I = 5.6 x 10⁻⁵ kg/m³

Work done to change the rotational rate

K.E = ¹/₂Iω²

K.E = ¹/₂(5.6 x 10⁻⁵)(60 - 40)²

K.E = ¹/₂(5.6 x 10⁻⁵)(20)²

K.E = 0.0112 J

Percentage decrease of rotational inertia when the outermost particle is removed

I₂ = md² + m(2d)²

I₂ = 5md²

ΔI = 14md² - 5md²

ΔI = 9md²

η = (ΔI/I) x 100%

η = (9md²/14md²) x 100%

η = 64.29 %

Learn more about rotational inertia here: brainly.com/question/14001220

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