The system below is a variation on Atwood\'s machine. It consists of two masses

Question

The system below is a variation on Atwood's machine. It consists of two masses connected by a massless inextendable cord. One of the masses sits on a revolving turntable and the other hangs below the turntable directly below its centre. The coefficient of static friction between the mass m1 and the turntable is mgw=0.931.If the mass of m1 is 26.00kg and the mass of m2 is 59.00kg, what is the minimum speed at which the turntable must spin, in order to prevent mass m1 from sliding?
m1 is placed at a distance of 0.290 m from the center of the turntable. Express your answer as an angular speed. Angular speed is expressed in rad/sec (rad/s) where there are pi rads in 180 degrees.

Explanation / Answer

let angular speed is w rad/sec

let tension in the string be T.

then as the system is not moving,

writing force balance equation for hanging mass m2:

T=m2*g

==>T=59*9.8=578.2 N

now, the mass m1 can slide both inward and outward.

and depending upon its motion, friction force's direction will also change. (as friction opposes motion)

if m1 tries to slide inward,

then friction will act outwards along with centripetal force.

then T=m1*w^2*r+friction force

similarly , if m1 tries to slide outward, friction will act inwards and force balance equation will become:

T+friction force=m1*w^2*r

from both the equation, it is clearly observed that, for minimum value of w, we need to use the first scenario where friction force is acting outward.

now, friction force=friction coefficient*normal force

=0.931*m1*g

=0.931*26*9.8

=237.22 N

then T=26*w^2*0.29+237.22

==>26*w^2*0.29=578.2-237.22=340.98

==>w=sqrt(340.98/(26*0.29))=6.7248 rad/sec

hence the minimum angular velocity is 6.7248 rad/sec at which the mass m1 wont slide.

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