Help needed - Spinning wheel, torque, angular acceleration.

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SUMMARY

The discussion centers on calculating the torque and linear force exerted on a spinning cylindrical wheel with a mass of 0.013 kg and an initial angular velocity of 78.54 rad/s, which decelerates under an unknown constant angular deceleration. The calculations yield an angular deceleration of -3477 rad/s², a moment of inertia of 4.844 x 10-7 kg·m², and a torque of -0.00956 N·m. The linear force exerted on the metal rod, which is inserted at a radius of 0.6 cm, is calculated to be 1.593 N. The discussion highlights the complexity introduced by the non-constant deceleration when the rod is inserted before the wheel stops.

PREREQUISITES
  • Understanding of rotational kinematics and equations
  • Knowledge of torque and moment of inertia calculations
  • Familiarity with the parallel axis theorem
  • Basic principles of angular motion and forces
NEXT STEPS
  • Study the application of the parallel axis theorem in rotational dynamics
  • Learn about non-constant angular deceleration scenarios
  • Explore advanced torque calculations in dynamic systems
  • Investigate the relationship between linear and angular motion
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Students in physics, mechanical engineers, and anyone involved in dynamics and rotational motion analysis will benefit from this discussion.

Fjolvar
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Homework Statement


There is a spinning cylindrical wheel of mass 0.013 kg at a given constant angular velocity ωo (78.54 rad/s). The wheel instantaneously begins to decelerate at an unknown constant deceleration α. From when it begins to decelerate, the wheel spins a distance of θ (0.887 radians) before a metal rod is placed inside of the wheel at a distance of 0.6cm radius from the origin, bringing the wheel to a stop by making contact with one of the spokes BEFORE the wheel completely decelerates (so the wheel is still in motion as the rod is placed inside). What is the torque and linear force exerted on the metal rod when stopping the wheel?

Homework Equations


θ = ωot + (1/2)α t2
ω2 = ωo2 + 2α (θ-θo)
I = (1/2)Mr2
L = r x p = r x mv = Iω
τ = I*α = r x F = r*F*sinθ

The Attempt at a Solution


Okay, so I chose to use rotational kinematic equations to determine the torque value.

First I calculated the angular deceleration by using the rotational kinematic equation, assuming ω=0 and θo=0 :
ω2 = ωo2 + 2α (θ-θo). This gave me α = -3477 rad/s2.

Then I calculated the moment of intertia of a cylinder using the parallel axis theorem about the point of contact, I = Icm + mr2, which gave me a value of 4.844*10-7.

Finally plugging I and α into the equation for torque τ = I*α, I got τ = -0.00956 N*m and force value of 1.593 N using the equation τ= r*F*sinθ (assuming θ to be 90 degrees since the rod is inserted perpindicular to the wheel).

I would greatly appreciate any feedback on my methods of solving this problem, since I'm unsure if this answer or method is correct. Thank you for your time.
 
Last edited:
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I just made the observation, since the rod is placed into the wheel before it fully decelerates to zero, this makes for non-constant deceleration and thus makes the rotational kinematic equations unuseable correct?
 

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