Q.

A spool has the shape shown in figure. Radii of inner and outer cylinders are R and 2 R respectively. Mass of the spool is 3 m and its moment of inertia about the shown axis is 2mR2. Light threads are tightly wrapped on both the cylindrical parts. The spool is placed on a rough surface with two masses m1 = m and m2 = 2m connected to the strings as shown. The string segment between spool and the pulleys P1 and P2 are horizontal. The centre of mass of the spool is at its geometrical centre. System is released from rest. What is minimum value of coefficient of friction between the spool and the table so that it does not slip?

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answer is 00000.11.

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Detailed Solution

Let acceleration of COM of the spool be a0 towards right, and acceleration of m2 be a2 downward and that of m1 be a1 upward f = friction, α = acceleration of the spool 2Rα=a0         .......(1)Rα+a0=a2      .....(2)2Rα+a0=a1    .......(3)T1−mg=ma1T1−mg=ma0+2Rα=2ma0     .....(4)   ∵Rα=a022mg−T2=2ma22mg−T2=2ma0+Rα=3ma0    ........(5)T2+f−T1=3ma0                             .........(6)T2R−T12R−f2R=lαT2−2T1−2f=2mRα ∵I=2mR2T2−2T1−2f=ma0                             .......(7)Solving, (4), (5), (6) & (7) we get  f=−mg3Negative sign means correct direction of friction is towards right side.∵f≤μN ∴mg3≤μ3mg ∴19≤μ
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