A solid cylinder of mass $50 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ is free to rotate about the…

A solid cylinder of mass $50 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ is free to rotate about the horizontal axis. Amassless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions $\mathrm{s}^{-2}$ is
  1. $25 \mathrm{~N}$
  2. $50 \mathrm{~N}$
  3. $78.5 \mathrm{~N}$
  4. $157 \mathrm{~N}$

Solution

Here $\alpha=2$ revolutions $/ \mathrm{s}^{2}=4 \pi \mathrm{rad} / \mathrm{s}^{2}$ (given)
$\begin{aligned} \mathrm{I}_{\text {cylinder }} &=\frac{1}{2} \mathrm{MR}^{2}=\frac{1}{2}(50)(0.5)^{2} \\ &=\frac{25}{4} \mathrm{Kg}-\mathrm{m}^{2} \\ \text { As } \tau=\mathrm{I} \alpha \text { so } \mathrm{TR}=\mathrm{I} \alpha \end{aligned}$
$\Rightarrow \mathrm{T}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{\left(\frac{25}{4}\right)(4 \pi)}{(0.5)} \mathrm{N}=50 \pi \mathrm{N}=157 \mathrm{~N}$ .

Asked in: JEE Mains - Rotational Motion - Test 4

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