Q.

A clockwise torque of 6 N-m is applied to the circular cylinder as shown in the figure. There is no friction between the cylinder and the block.

Moderate

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By Expert Faculty of Sri Chaitanya

The cylinder will be slipping but the system does not move forward.

The system cannot move forward for any torque applied to the cylinder.

The acceleration of the system will be 1m/s2 forward.

The angular acceleration of the cylinder is 10 rad s-2.

answer is 3,4.

(Detailed Solution Below)

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Detailed Solution

a=f1−f2m1+m2=0.4×6g−0.5×3g6+3⇒a=1ms−2α=τapplied −τfriction I=6−f1R126(R)2=6−0.4×6g(0.2)12×6×(0.2)2=10rads−2

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