Question

A stone of mass $1kg$ is tied to a string $2m$ long and it’s rotated at constant speed of $40m{s}^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g=10m{s}^{-2}$]

• A. $\frac{81}{79}$
• B. $\frac{79}{81}$
• C. $\frac{19}{12}$
• D. $\frac{12}{19}$

Answer 1

Answer: (B)

Free body diagram (FBD) of a stone moving in a vertical circular path, which has tension force at point A and $B$ as represented by $T_A$ and $T_B$ respectively as given below in the figure,

Given, mass of stone $(m)=1kg$,
length of the string $(R)=2m$ and
rotating linear speed $(v)=40m{s}^{-1}$
As, we know that the tension at position $A$,
$T_A=\frac{m{v}^2}{R}+mg$
$\left(\therefore F_c=\frac{m{v}^2}{R}\right)$
$\Rightarrow T_A=\frac{1\times (40{)}^2}{2}+1\times 10=810N$
Similarly, tension at position $B$
$\Rightarrow T_B=\frac{m{v}^2}{R}-mg=\frac{1\times (40{)}^2}{2}-1\times 10=790N$
So, the ratio of $T_B$ and $T_A$ i.e.,
$\frac{T_B}{T_A}=\frac{790}{810}=\frac{79}{81}$
Hence, the ratio of tension at position $B$ and tension at position $A$ is $79:81$.