**Motion of mass m:**There are two forces acting on mass

**m**,

**i)**the force of gravity

**mg**and

**ii)**the tension in the string

**T1**. Let its acceleration be

**a1**. So from Newton's laws we have,

**Motion of mass M:**There are four forces acting on mass

**M**,

**i)**the force of gravity

**Mg**,

**ii)**the tension in the string connecting to mass

**m**

**i.e.**

**T2**,

**iii)**the forces on the two strings holding mass

**m**of

**T1**each. Let the acceleration of mass

**M**be

**a2**. So we have,

Since the force of gravity

**Mg**passes through the axis of rotation of mass

**M**, it does not produce any torque in it. However, the rest of the three forces produce torques inducing rotation in the mass. Since the tensions

**T2**act at a distance of

**R**from the axis, they produce a torque of

**T2R**each in the clockwise direction. The tension

**T1**acts at a distance of

**2R**from the axis of rotation of mass

**M**and so it induces a torque of

**T1(2R)**on the mass also in the clockwise direction. Let the angular acceleration of mass

**M**be . Then we have,

Since the threads holding mass

**M**do not slip we have,As seen in the figure, suppose that the mass

**M**rotates an angle , each of the two threads unwrap a length of and so the mass**M**falls down this distance. Meanwhile, the rotation of mass**M**also releases a length of of the thread holding mass**m**, making the distance between mass**M**and mass**m**to increase by this additional distance. Consequently the total distance moved by mass**m**is as shown in the figure. From this discussion and**(4)**it is clear that,Now using

**(1), (2), (5)**and**(6)**we can solve for**a1**as follows,
## 3 comments:

PLEASE EXPLAIN THE USE OF 2THETA R

The thicker part of the spindle has a radius of 2R, so for the same angular rotation theta, 2(theta)R rolls out.

the solutions are really helpful in cases of such tough problems

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