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Regarding Angular Velocity:

  • Thread starter Zhadows
  • Start date
  • #1
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Homework Statement



A 5.0 kg, 60-cm diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.

a. What is the cylinder's initial angular Acceleration?
b*. What is the cylinder's angular Velocity when it is directly below the axle?


Homework Equations



I=Icm+MD^2

A=T/I

The Attempt at a Solution



Part A: I utilized the Parallel axis theorem because it was rotating off the center of mass.
I=Icm+MD^2->[I of disc] (1/2 (5)(0.3)^2) + 5(.3)^2= 0.675

For the torque: T=R*F->(0.3)(5*9.8)=14.7

14.7/0.675= 21.78 Rad/S^2

Can someone tell me if I did something incorrect? Also:

Part B: I have no idea where to start. This is where I really need help.

Thanks.
 

Answers and Replies

  • #2
20
0
Try Conservation of rotational kinetic energy.
 
  • #3
Try using the kinematics equations replacing x with (radians), v with (angular velocity) and a with angular acceleration. I will reply if with additional information if you follow up on this.
 
  • #4
alphysicist
Homework Helper
2,238
1
I think conservation of total energy is the best approach here; the increase in rotational kinetic energy must equal the decrease in gravitational potential energy. A kinematic approach would be more difficult since the angular acceleration is not constant.
 

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