A solid, horizontal cylinder of mass 13.4 kg and radius 1.00 m rotates with an angular speed of 7.50 rad/s about a fixed vertical axis through its center. A 0.250-kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.   rad/s=

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A solid, horizontal cylinder of mass 13.4 kg and radius 1.00 m rotates with an angular speed of 7.50 rad/s about a fixed vertical axis through its center. A 0.250-kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system. 
 rad/s=

 

Expert Solution
Step 1

Given: The mass of horizontal cylinder is 13.4 kg. 

           The radius is 1.00 m. 

           The angular speed of cylinder is 7.50 rad/s.

           The mass of the putty is 0.250 kg. 

           The distance between the putty and the cylinder is 0.900 m. 

To determine: The final angular speed of the system. 

The angular momentum of the system is conserved, that is the initial angular momentum is equal to final angular momentum of the system. 

The formula to calculate initial angular momentum is 

Li=Iw ....................... (1) 

where I is the momentum of inertia of the cylinder and w is the angular frequency. 

The formula to calculate the moment of inertia is 

I=12mr2

Substitute 12mr2 for I in equation (1) 

Li=12mr2w

Substitute 13.4 kg for m, 1.00 m for r and 7.50 rad/s for w in the above equation, 

Li=12×13.4 kg×1.00 m2×7.50 rad/s=50.25 kg.m2/s

The final angular momentum is 

Lf=Icylinderωf+Iputtyωf

where Iputty  is mr2

Lf=12mr2ωf+mputtyrp2ωf

 

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