Question

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

• 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 $(m)=1 kg$,
length of the string $(R)=2 m$ and
rotating linear speed $(v)=40 ms ^{-1}$
As, we know that the tension at position $A$,
$T_{A}=\frac{m v^{2}}{R}+m g$
$\left(\therefore F_{c}=\frac{m v^{2}}{R}\right)$
$\Rightarrow T_{A} =\frac{1 \times(40)^{2}}{2}+1 \times 10=810 N$
Similarly, tension at position $B$
$\Rightarrow T_{B}=\frac{m v^{2}}{R}-m g=\frac{1 \times(40)^{2}}{2}-1 \times 10=790 N$
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 $79: 81$.