Question

A solid cylinder of mass $50kg$ and radius $0.5m$ is free to rotate about the horizontal axis. A massless string is wound end round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2rev/s^2$

• A. $25N$
• B. $50N$
• C. $78.5N$
• D. $157N$

Answer 1

Answer: (D)

$157N$

Mass, $m=50$kg

Radius, $r=0.5$m,

Angular acceleration $\alpha =2rev/s^2=4\pi red/s^2$

Torque produces by the tension a string$=T\times R=T\times 0.5=T/2$ N/m ....$\left(1\right)$

we know,

$T=I_{\alpha }$ ........$\left(2\right)$

From equation $\left(1\right)$ & $\left(2\right)$

$\frac{T}{2}=I\alpha =\left(\frac{M{R}^2}{2}\right)\times \left(4\pi \right)\frac{rad}{s^2}$

$\therefore$ $T_{solidcylinder}=\frac{M{R}^2}{2}$

$\frac{I}{2}=\frac{50\times {0.5}^2}{2}\times 4\pi$

$T=50\pi$

$T=157N$.