Two particles of equal mass m are connected t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the distance from $B$ to the first particle be $r$ and the distance between the two particles be $2r$. Let $\omega$ be the angular velocity of

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(A) Let the distance from $B$ to the first particle be $r$ and the distance between the two particles be $2r$. Let $\omega$ be the angular velocity of the system.The tension in the outer part of the rope (between the two particles) is $T_2$. This tension provides the centripetal force for the particle at end $A$:$T_2 = m \omega^2 (r + 2r) = 3m \omega^2 r$The tension in the inner part of the rope (between $B$ and the middle particle) is $T_1$. This tension provides the centripetal force for both particles:$T_1 = T_2 + m \omega^2 r$$T_1 = 3m \omega^2 r + m \omega^2 r = 4m \omega^2 r$The ratio of the tension in the smaller part $(T_1)$ to the tension in the larger part $(T_2)$ is:$\frac{T_1}{T_2} = \frac{4m \omega^2 r}{3m \omega^2 r} = \frac{4}{3}$

Similar Questions

$A$ freely falling body has attained a velocity of $2 \,m/s$. If the weight of the body is now opposed by a constant upward air resistance force equal to its weight, what is the total distance travelled by the body before it comes to rest (in $\,m$)?

What is the minimum acceleration with which a person can slide down a rope,if the breaking strength of the rope is $\frac{2}{3}$ of the person's weight?

$A$ mass $1\, kg$ is suspended by a thread. It is
$(i)$ lifted up with an acceleration $4.9\, m/s^2$
$(ii)$ lowered with an acceleration $4.9\, m/s^2$.
The ratio of the tensions is

Two masses of $4\, kg$ and $5 \,kg$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of the $5\, kg$ mass is ........... $m/s^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module