Question

Two rocks are tied to massless strings and whirled in nearly horizontal circles so that the time to travel around the circle once is the same for both. One string is twice as long as the other. The tension in the longer string is twice the tension in the shorter one. What is the mass m of the rock at the end of the shorter string compared to the mass m2 of the rock at the end of the longer one? A. m1 = m2/4 B. m1=m2 C. m m2 D. m1=2m2 E. m1= 4m2

Answer 1

Answer:$m_1=m_2$

Explanation:

Given

Time period for both string is same

$\frac{2\pi r}{v_1}=\frac{2\pi 2r}{v_2}$

$2v_1=v_2$

and tension in string 2 is  twice the first string

$2T_1=T_2$

Tension will provide centripetal acceleration

$2\frac{m_1v_1^2}{r}=\frac{m_2v_2^2}{2r}$

$2\frac{m_1v_1^2}{r}=\frac{m_2\times 4v_1}{2r}$

thus $m_1=m_2$