Question

Two masses M and m are attached to a vertical axis by weightless threads of combined length $l$. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω. If the tensions in the threads are the same during motion , the distance of M from the axis is :
A. $\frac{Ml}{M+m}$
B. $\frac{ml}{M+m}$
C. $\frac{M+m}{M}l$
D. $\frac{M+m}{m}l$

Answer 1

Hint:Tension is a type of force that works in medium lengths , especially those that are flexible , such as rope or cord. Tension force remains a gravitational force.Tension by the force between these masses M and m will be equal as it is acting upon the same length of rope . Hence the individual tensions from the respective masses can be calculated and then can be equated to each other from the axis to get the desired value of the distance from the axis.



Formula used:



Complete answer:
Assuming the two masses are attached to a weightless threads in a vertical axis which are of some distance from each other. Let consider a point in between of the two masses from which the axis of rotational motion is done to be represented by O. Now as stated in the question it is making a rotation with an angular velocity of ω . Angular velocity is a velocity which describes the rate of change of angular position of a rotating body . Now let us take the distance between the mass M and the axis of rotation be x units. Also assume the tension in the thread be T. Then from the formula of tension
We have ,
Tension(T)=mass of the body ×distance from the axis of rotation×${{\omega }^{2}}$
Now it is given that the comvined length of the thread from the mass M to m is $l$.Since the
Distance between the axis of rotation and mass M body = x units
Then distance between the axis of rotation and mass m body= $l$-x units [∵ d(M)+d(m)= $l$]
∴ Tension due to mass M (T)= Mx${{\omega }^{2}}$
And tension due to mass m , T=m($l$-x) ${{\omega }^{2}}$
Since, Tension by the force between these masses M and m will be equal as it is acting upon the same length of rope . Hence the individual tensions from the respective masses can be calculated and then can be equated to each other from the axis to get the desired value of the distance from the axis.
This implies that, Mx${{\omega }^{2}}$=m($l$-x) ${{\omega }^{2}}$
               $\Rightarrow $ Mx${{\omega }^{2}}$=m$l$${{\omega }^{2}}$- mx${{\omega }^{2}}$
              $\Rightarrow $ Mx${{\omega }^{2}}$- mx${{\omega }^{2}}$=m$l$${{\omega }^{2}}$
              $\Rightarrow $ x(M${{\omega }^{2}}$- m${{\omega }^{2}}$)= m$l$${{\omega }^{2}}$
              $\Rightarrow $x= $\frac{ml}{M+m}$units
Thus, the correct option is option B.






Note:Tension from the same type of material is same only even after the different set of origin in a considered single frame of objects as it became balanced to each other . The formula of the tension should be remembered properly as it became quite different from question to question according to the situations and data provided in the question.