Calculating Acceleration of a Cylinder on an Incline

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SUMMARY

The discussion focuses on calculating the acceleration of a solid cylinder rolling down an incline. The cylinder has a radius of 10 cm and a mass of 12 kg, rolling a distance of 6.0 m on a 30-degree incline. The established formula for the acceleration is a = (2gsinθ)/3, derived from the moment of inertia I = (1/2)mr² and the acceleration of a block down a frictionless incline, which is gsinθ. The factor of two-thirds arises from the relationship between translational and rotational motion in rolling objects.

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  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia
  • Basic knowledge of trigonometry, specifically sine functions
  • Concept of rolling without slipping
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  • Learn about the relationship between linear and angular acceleration
  • Explore the concept of moment of inertia for different shapes
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The first part to an exam question I had to sit at the end of last year. It defeated me then, and I still can't do it now, which worries me more than a little!

Homework Statement


A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls
without slipping a distance L = 6:0 m down a roof that is inclined at an
angle [tex]\theta[/tex] = 30[tex]\circ[/tex]
Show that the acceleration of the cylinder is given by
a= [tex]\frac{2gsin\theta}{3}[/tex]


Homework Equations


The acceleration of a block down a frictionless incline is just gsin [tex]\theta[/tex]
The moment of inertia of a solid cylinder of mass m and radius r
is given by I = [tex]\frac{1}{2}[/tex]mr[tex]^{2}[/tex]


The Attempt at a Solution


As 2, but I really don't know where the factor of two thirds comes from. The question was worth 6 marks.
 
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Fantastic. Thanks :smile:
 

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