Question

A thin string of negligible mass with one end fixed to the roof is wound around a circular disc of radius 2 m and mass, 10 kg as shown below. The disc rolls vertically down under the action of its own weight.Considering acceleration due to gravity as 10 m/s${}^2$, the tension is the string is

• A. 0 N
• B. 25.0 N
• C. 33.3 N
• D. 50 N

Answer 1

The correct option is C 33.3 N

$\text{Torque}=I\alpha =\frac{m{r}^2}{2}\times \frac{a}{r}\Rightarrow T=\frac{amr}{2}$

$\text{Tension,}p=\frac{T}{r}=\frac{ma}{2}$

By applying D'alembert's principle

$ma=mg-p=mg-\frac{ma}{2}$

$\frac{3ma}{2}=mg$

$\therefore a=\frac{2}{3}g=\frac{20}{3}m/se{c}^2$


$\text{Tension,}P=\frac{T}{r}=\frac{ma}{2}=\frac{10}{2}\times \frac{20}{3}=\frac{100}{3}N$

$P=33.3N$