Consider a 20kg uniform circular disk of radius 0 . 2m . It is pin...

Consider a $20kg$ uniform circular disk of radius $0 . 2m$ . It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force $F = 20N$ through a massless string wrapped around its periphery as shown in the figure.

Consider a 20kg uniform circular disk of radius 0 . 2m . It is pin supported at its center and is at rest initially.

Suppose the disk makes $n$ number of revolutions to attain an angular speed of $50rad{s}^{- 1}$ . The value of $n$ , to the nearest integer, is _______ .

[Given : In one complete revolution, the disk rotates by $6 . 28rad$ ]

Correct answer

Step-by-step solution

$\alpha = \dfrac{τ}{I} = \dfrac{F . R .}{m{R}^{2} / 2} = \dfrac{2F}{mR}$

$\alpha = \dfrac{2 \times 200}{20 \times \left(0 . 2\right)} = 10rad / {s}^{2}$

${\omega}^{2} = {\omega}_{0}^{2} + 2\alpha\Delta\theta$

${\left(50\right)}^{2} = {0}^{2} + 2\left(10\right)\Delta\theta ⇒ \Delta\theta = \dfrac{2500}{20}$
$\Delta\theta = 125rad$

No. of $= \dfrac{125}{2\pi} \approx 20$ revolution

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