Question

A disc rotating about an axis experiences an angular retardation proportional to the angle with which it rotates. If the rotational kinetic energy reduces of an amount ∆E when it rotates through an angle theta then ∆E is proportional to​

Answer 1

Answer:

E is directly proportional to 3/2 power of theta

Answer 2

The reduce in rotational kinetic energy, ΔE is proportional to θ².

A disc rotating about an axis experiences an angular retardation proportional to the angle with which it rotates.

If the rotational kinetic energy reduces of an amount ΔE when it rotates through an angle θ , then ΔE is proportional to ..

We know, The magnitude of angular acceleration/retardation is the rate of change of angular velocity.

i.e., α = dω/dt = (dω/dθ) × (dθ/dt) = (dω/dθ) × ω = ω(dω/dθ)

[ ∵ rate of change of angular displacement is known as angular velocity. i.e., ω = dθ/dt ]

Given, angular retardation is proportional to the angle which it rotates.

∴ α ∝ -θ [ negative sign indicates that it is retardation ]

⇒ ω(dω/dθ) ∝ -θ

⇒ ω(dω/dθ) = -Cθ [ where C is proportionality constant ]

$\implies\int\limits^{\omega_f}_{\omega_i}\omega d\omega=-C\int\limits^{\theta}_0\theta d\theta$

$\implies\frac{1}{2}[\omega_f^2-\omega_i^2]=-C\frac{\theta^2}{2}\\\\\implies\frac{1}{2}I[\omega_i^2-\omega_f^2]=CI\frac{\theta^2}{2}$

$\implies\Delta E=CI\frac{\theta^2}{2}$

$\implies\Delta E\propto\theta^2$

Here it is clear that reduce in rotation kinetic energy is proportional to the square of the angle with which it rotates.

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