Acceleration of an object rollling down a incline.

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SUMMARY

The acceleration of a cylinder rolling down an incline is determined using the formula a = (2/3)g sin(Θ), where g is the acceleration due to gravity and Θ is the incline angle. The derivation involves applying the torque equation ∑τ = Iα and the parallel-axis theorem to find the moment of inertia. The torque acting on the cylinder is given by (Mg sin(Θ))R, which is equated to the rotational inertia (1/2MR² + MR²)(a/R). Understanding the relationship between linear and angular acceleration is crucial for this derivation.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with the parallel-axis theorem
  • Knowledge of moment of inertia for solid cylinders
  • Basic principles of rolling motion without slipping
NEXT STEPS
  • Study the derivation of the parallel-axis theorem in detail
  • Learn about the moment of inertia for various shapes
  • Explore the relationship between linear and angular acceleration in rolling objects
  • Investigate real-world applications of rolling motion in physics
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Students studying classical mechanics, physics educators, and anyone interested in understanding the dynamics of rolling objects on inclined planes.

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Homework Statement


A cylinder of mass M and radius R rolls(without slipping) down an inclined plane whose incline angle with the horizontal is Θ. Determine the acceleration of the cylinder's venter of mass.(Derive the formula)


Homework Equations


I have the answer, but I just don't understand:

∑τ=Iα, use the parallel-axis theorem and a=Rα.
(Mg sin Θ)R=(1/2MR²+MR²)(a/R)
Rg sin Θ=3/2Ra
a=2/3g sinΘ

The Attempt at a Solution


Here's my question. Why do we need to use the parallel-axis theorem? isn't that the torque is just 1/2MR² ? Maybe I am confused with the concept of Torque. I don't know...Please help!
 
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When the cylinder is rolling without slipping, the relative motion of point of contact is zero.
So it acts as the axis of rotation of the center of mass. That is why you have to find the moment of inertia about this axis by using parallel-axis theorem.
 

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