A uniform cylinder of mass $m$ and radius $r$ is rolling down a slope of inclination $\theta$. The cylinder rolls without slipping. You may take the acceleration due to gravity to be $g$. At what rate does the cylinder accelerate down the slope?

The answer is $\frac23 g \sin\theta$

How do you get to this answer?


closed as off-topic by John Rennie, akhmeteli, Emilio Pisanty, Qmechanic Oct 27 '13 at 0:15

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  • $\begingroup$ You could get the answer by using the conservation of energy & Newton's laws. $\endgroup$ – Kyle Kanos Oct 25 '13 at 19:25
  • 1
    $\begingroup$ possible duplicate of Cylinder rolling down an inclined plane $\endgroup$ – Ali Oct 25 '13 at 20:05
  • $\begingroup$ You get the answer by trying to do it, then posting your work here so people can guide you, instead of fishing for ready-made answers $\endgroup$ – Pranav Hosangadi Oct 26 '13 at 2:37

Gravitational force $\vec{F_g}$ is trying to rotate the cylinder around point A.
Lever arm has length $r_c= r\,\sin\theta$
Torque $\tau$ is $$\tau = F_g \, r_c = m \, g\, r \, \sin\theta$$ Torque causes angular acceleration $\alpha$ of the cylinder.
We are interested in acceleration $\vec{a}$ of point C. cylinder on a slope Absolute value of point C acceleration is $|\vec{a}| = \alpha \, r$.
We need to compute angular acceleration $\alpha$ which is quotient of torque $\tau$ and moment of inertia $I$. $$\alpha = \frac{\tau}{I}$$

Moment of inertia of a cylinder is $I_{cylinder}=\frac{1}{2}\,m\,r^2$. Axis of rotation doesn't intersect center of mass, we will use parallel axis theorem. $$I = I_{cylinder} + m \, r^2 = \frac{1}{2}\,m\,r^2 + m\,r^2 = \frac{3}{2}\,m\,r^2$$

And from moment of inertia we compute acceleration of point C $$a=\alpha\,r=\frac{\tau}{I}\,r=\frac{m\,g\,r\,\sin\theta}{\frac{3}{2}\,m\,r^2}\,r=\frac{2}{3}\,\frac{m\,g\,r^2\,\sin\theta}{m\,r^2}$$

And finally $$a=\frac{2}{3}\,g\,\sin\theta$$

  • $\begingroup$ Nice sketch, is it GeoGebra or what? $\endgroup$ – ja72 Oct 26 '13 at 1:23
  • $\begingroup$ @ja72: yes it is GeoGebra, nice tool $\endgroup$ – user1086737 Oct 26 '13 at 9:17

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