The value of force is 211N. Thus, the cylinder undergoes through 6 rotations with a force is 211N
A cylinder of mass 5 kg and radius 50cm is rotating at angular speed of 5 rad/sec.
To find the force that must be applied tangentially to the cylinder to bring it to rest in 6 revolutions
As per the given criteria
Rotation if the wheel = 5 rad/ sec
Now, total rev = 6
We know
0 = (10/6) ² - 2 lamda × 6
= 100/36=12 lamba
Lamda= 33.33 rev/sec
Lamda = 2π×33.33 rad/sec²
Now the moment of inertia
I=1/2mr²
=1/2 (0.5) ²× 5
=0.625 kgm²
Since the force F, now torque
F=F.r=0.62 F
Now, torque = I × lamda
= 0.625×33.33×2π
Or 0.62 F= 130.82
F= 211 N
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