Dynamics of Rotational Motion and hinge

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SUMMARY

The discussion focuses on the dynamics of rotational motion involving a thin, uniform rod of length 0.8 m and mass 2 kg, connected to a frictionless hinge. Key calculations include angular velocity and angular acceleration at various angles: θ=90°, θ=60°, and θ=0°. The correct results identified are 0 rad/s for angular velocity at θ=90°, 74 rad/s² for angular acceleration at θ=90°, 8.6 rad/s for angular velocity at θ=60°, and 12 rad/s for angular velocity at θ=0°. The participants emphasize the importance of using the correct expressions for angular acceleration and energy conservation principles.

PREREQUISITES
  • Understanding of rotational dynamics
  • Familiarity with torque and moment of inertia
  • Knowledge of energy conservation in rotational systems
  • Ability to apply angular kinematics equations
NEXT STEPS
  • Study the derivation of the moment of inertia for a uniform rod
  • Learn about torque calculations in rotational motion
  • Explore energy conservation principles in rotational dynamics
  • Practice solving problems involving angular acceleration and velocity
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Physics students, educators, and anyone interested in understanding the principles of rotational motion and dynamics, particularly in mechanical systems.

denny2
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One end of a thin, uniform rod is connected to a frictionless hinge as shown in Figure 1. The rod has a length of 0.8 mand a mass of 2 kg. It is held up in the horizontal position (θ=90∘) and then released.
1)Calculate the angular velocity of the rod at θ=90∘.
2)Calculate the angular acceleration of the rod at θ=90∘.
3)Calculate the angular velocity of the rod at θ=60∘
4)Calculate the angular acceleration of the rod at θ=60∘
5)Calculate the angular velocity of the rod at θ=0∘.
6)Calculate the angular acceleration of the rod at θ=0∘.
1067.PNG

Homework Equations


mgh=1/2 Iw^2
a = ((mgcos)1/2)/I

The Attempt at a Solution


I got these answers and only 1 and 6 were correct
1)0 rad/s
2)74 rad/s^2
3)8.6 rad/s
4)64 rad/s^2
5)12 rad/s
6)0 rad/s^2

Any help is appreciated thank you :)
 
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denny2 said:
Calculate the angular acceleration of the rod at θ=90∘.
3)Calculate the angular velocity of the rod at θ=60∘
4)Calculate the angular acceleration of the rod at θ=60∘
5)Calculate the angular velocity of the rod at θ=0∘.

you are having problem with above - so check expression for angular acceleration.
moment of inertia xangular acceleration = torque on the rod =forcex the the perpendicular distance from the axis of rotation
and for angular velocity one can use the work- energy conversion
work done by a torque is torquex angular displacement= rotational energy
 
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denny2 said:
One end of a thin, uniform rod is connected to a frictionless hinge as shown in Figure 1. The rod has a length of 0.8 mand a mass of 2 kg. It is held up in the horizontal position (θ=90∘) and then released.
1)Calculate the angular velocity of the rod at θ=90∘.
2)Calculate the angular acceleration of the rod at θ=90∘.
3)Calculate the angular velocity of the rod at θ=60∘
4)Calculate the angular acceleration of the rod at θ=60∘
5)Calculate the angular velocity of the rod at θ=0∘.
6)Calculate the angular acceleration of the rod at θ=0∘.
1067.PNG

Homework Equations


mgh=1/2 Iw^2
a = ((mgcos)1/2)/I

The Attempt at a Solution


I got these answers and only 1 and 6 were correct
1)0 rad/s
2)74 rad/s^2
3)8.6 rad/s
4)64 rad/s^2
5)12 rad/s
6)0 rad/s^2

Any help is appreciated thank you :)
We cannot tell where you are going wrong if you do not post your working.
 

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