Question

A particle of mass m1 is fastened to one end of a string. Another particle of mass m2 is fastened to the middle point of that string. The other end of the string is fastened to a fixed point on a smooth horizontal table. The particles are then projected so that the two portions of the string are always in the same straight line and describe horizontal circles. Find the ratio of tensions in the two parts of the string.

a) $\sf\dfrac{m_1}{m_1+m_2}$

b) $\sf\dfrac{m_1+m_2}{m_1}$

c) $\sf\dfrac{2m_1+m_2}{2m_1}$

d) $\sf\dfrac{2m_1}{m_1+m_2}$​

Answer 1

Laws of motion

This question says us that the mass $m_1$ and mass $m_2$ are connected to the string and the centre is $o$ and these are moving in a horizontal circle. So when they move in a horizontal circle, the angular velocity $\omega$ of both will be the same and here there will be some tension, the tension $t_1$ will be in the mass $m_1$ and the tension $t_2$ will be in the mass $m_2$. We have to calculate the ratio of tensions in the two parts of the string.

[Please refer to the attachment for a better clarity]

Let us consider that $\omega$ be the angular velocity.

$r$ is the radius from $o$ to $m_2$ then $2r$ will be the radius from $o$ to $m_1$.

Tension due to mass $m_1$;

$\implies T_1 = T_2 + m_2 \omega^2 r$

Tension due to mass $m_2$;

$\implies T_2 = m_1 \omega^2 2r$

Substituting the value of $T_2$ in $T_1$

$\implies T_1 = m_1 \omega^2 2r + m_2 \omega^2 r \\ \\ \implies T_1 = \omega^2 r(2m_1 + m_2)$

We have to calculate the ratio of tensions in the two parts of the string. Therefore,

$\implies \dfrac{T_1}{T_2} = \dfrac{\omega^2 r (2m_1 +m_2)}{m_1 \omega^2 2r} \\ \\ \implies \dfrac{T_1}{T_2} = \dfrac{\cancel{\omega^2 r} (2m_1 +m_2)}{2m_1 \cancel{\omega^2 r}} \\ \\ \implies \dfrac{T_1}{T_2} =\dfrac{2m_1 + m_2}{2m_1}$

Hence, the the ratio of tensions in the two parts of the string is $\dfrac{2m_1}{m_1+m_2}$.

Answer 2

Answer:

c) (2m1 + m2)/(2m1)

Explanation:

A particle of mass m1 is fastened to one end of a string.

Let's say the angular speed is ω and radius is R & R/2 is the radius of inner circle.

Tension (T1) in other end of the string = m1 × Rω² ------- (eq 1)

Now,

Tension due to mass (m2) = T1 + m2 × r/2 × ω²

Substitute value of T1 from (eq 1) in T2

T2 = m1 × Rω² + m2 × R/2 × ω²

T2 = Rω² × (m1 + m2/2)

We need to find out the ratio of tensions in the two parts of the string.

So,

T2/T1 = [Rω² (m1 + m2/2)/(m1 × R × ω²)

T2/T1 = [Rω² {(2m1 + m2)/2})/(m1 × R × ω²)

T2/T1 = Rω² [(2m1 + m2)/2]/(m1 × Rω²)

Rω² throughout cancel, we left with

T2/T1 = (2m1 + m2)/2m1

Therefore, the ratio of tensions in two parts of the string is (2m1 + m2)/2m1.