Two identical masses are connected by a (massless, stretchless) string across a

Question

Two identical masses are connected by a (massless, stretchless) string across a (massless, frictionless)
pulley. The mass on the right is held so that the string connected to it is horizontal. Everything is initially at rest; the tension is the string is zero. The mass on the right is then released and begins to swing downward. At what angle ? will the mass on the inclined plane begin to slide if the coefficient of static friction between the mass and the plane is 0.50? The indicated angle of the incline is 25.0O.

Explanation / Answer

Friction force F = *Normal force = *mgCos25

To start moving, Tension in string T > *mgCos25 + mgSin25

For swinging mass, Torque = mg*lCos where l = radius of swinging mass

Also Inertia I = ml^2

Using Torque = I we get, angular acceleration = mg*lCos/ml^2 = g/l *Cos

Putting = d/dt = (d/dt)(d/d) = d/d

d = g/l*Cos d

Integrating both the sides,

(^2)/2 = g/l*Sin + c

At = 0 we have = 0. Thus c = 0.

So, (^2)/2 = g/l*Sin

Tension in string T = centrifugal force = ml^2 = 2mg Sin

To start moving, Tension in string 2mg Sin > *mgCos25 + mgSin25

or Sin > (Cos25 + Sin25)/2

Sin > (0.5*Cos25 + Sin25)/2

Sin > 0.438

> 26 deg

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