Question

Two students attach two masses $m_1$ and $m_2$ to two different strings. The length of the string attached to $m_1$ is three-quarters the length of the string attached to $m_2$.
If both masses are swung in horizontal circles with equal constant speeds, the tensions in the strings are the same. What is the ratio $\frac{m_2}{m_1}$?

• A. $2$
• B. $4$
• C. $3/4$
• D. $4/3$
• E. $1/2$

Answer 1

The correct option is D $4/3$

As both the masses are moving with equal constant speeds therefore tangential accelerations of both masses are zero .

Since tensions in both strings are the same therefore centripetal forces on both the masses would also be equal because centripetal force is provided by tension in the any of the strings ,

centripetal force on mass $m_1$ = centripetal force on mass $m_2$

$\frac{m_1{v}_1^2}{r_1}=\frac{m_2{v}_2^2}{r_2}$

given $v_1=v_2=v,r_1=3/4r_2$

therefore ,$\frac{m_1{v}^2}{3/4r_2}=\frac{m_2{v}^2}{r_2}$

or $m_2/m_1=4/3$