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6. a 5 kg cylinder with a 50 cm radius is attached to the wall by an axis through its center and set into motion with an angular speed of 5 rad/sec. a force f is then applied to the edge of the cylinder as shown. the cylinder completes 6 full rotations before coming to a stop. what is the value of the force f?

Sagot :

The value of force is 211N.  Thus, the cylinder undergoes through 6 rotations with a force is 211N

How to calculate force of a cylinder?

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

Calculation :

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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