Question 307: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

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 $(m)=1 \mathrm{~kg}$,
length of the string $(R)=2 \mathrm{~m}$
and rotating linear speed $(v)=40 \mathrm{~ms}^{-1}$
As, we know that the tension at position $A$,
$$\begin{aligned}
& T_A=\frac{m v^2}{R}+m g \quad\left(	herefore F_c=\frac{m v^2}{R}ight) \
& \Rightarrow \quad T_A=\frac{1 	imes(40)^2}{2}+1 	imes 10=810 \mathrm{~N}
\end{aligned}$$
Similarly, tension at position $B$,
$$\Rightarrow \quad T_B=\frac{m v^2}{R}-m g=\frac{1 	imes(40)^2}{2}-1 	imes 10=790 \mathrm{~N}$$
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 $79: 81$.

Answer

$\frac{81}{79}$

Answer

$\frac{79}{81}$

Answer

$\frac{19}{12}$

Answer

$\frac{12}{19}$

Work, Energy and Power & Circular Motion

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

Choose the correct answer: