A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 Nm for 40 s, the speed increases to 21000 rpm. The diameter of the disk is __________ m. [2025]

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Solution

Q.
A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 $π$ Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is __________ m. [2025]

Ans.
(40)

Given m = 1 kg

$ω_{i} = 1800 rpm = 1800 \times \frac{2 π}{60} = 60 π \frac{rad}{\sec}$

$ω_{f} = 2100 rpm = 2100 \times \frac{2 π}{60} = 70 π \frac{rad}{\sec}$

$τ_{ex t} = 25 π N m, t = 40 \sec$

Using equation of motion, $ω_{f} = ω_{i} + α t$

$70 π = 60 π + α (40) \Rightarrow α = \frac{π}{4} rad / \sec^{2}$

Also, $τ = I α \Rightarrow τ = \frac{m R^{2}}{4} α$

$25 π = \frac{1 \times R^{2}}{4} \times \frac{π}{4} \Rightarrow R = 20 m$

Hence, diameter of disk = 2R = 2 $\times$ 20 = 40 m

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