Rolling Cylinders

Classical Mechanics Level 4

A uniform cylinder of mass $\SI[per-mode=symbol]{2}{\kilo\gram}$ and radius $\SI[per-mode=symbol]{0.5}{\meter}$ rests on a plank of mass $\SI[per-mode=symbol]{12}{\kilo\gram}$ . A force of $\SI[per-mode=symbol]{12}{\newton}$ is applied to the plank.

There is no friction between the ground and the plank, and the cylinder rolls without slipping on the plank.

If the acceleration of the plank in $\text{m/s}^2$ can be represented as $\frac ab,$ where $a$ and $b$ are coprime positive integers, what is the sum $a + b?$

28 37 46 69

1 solution

William G.
Jul 10, 2017

Remember that there is still friction between the block and the cylinder. Friction acts in the same direction as force applied on the cylinder and opposite to force applied for the block. In the block frame of reference, use the inertial force on the cylinder and friction for force analysis. Let M be the mass of the plank and m be the mass of the cylinder.

F (inertial) - F(friction) = m * a(cylinder)

m * a (plank) - F(friction) = m * a(cylinder) <======== Equation 1

F(friction) * r = 0.5 * m * r^2 (a/r) a = acceleration of cylinder from non-slipping identity

F(friction) = 0.5 * m * a(cylinder)

2 * F(friction) = m * a(cylinder) <============= Equation 2 into Equation 1

m * a(plank) - F(friction) = 2 *F(friction)

3 * F(friction) = m * a(plank)

F(friction) = m * a(plank) /3

Force analysis on plank

F(applied) - F(friction) = M * a(plank)

12 - m * a(plank) /3 = M * a(plank)

12 = a(plank) * (m/3 + M)

a(plank) = 12 / (m/3 + M)

a(plank) = 12/(2/3 + 12)

= 12/(38/3)

=36/38

=18/19

The two coprime numbers are a = 18 and b = 19, so the sum a + b = 37

I find it difficult to understand plz need help

Aman Bhandare - 3 years, 10 months ago

Rolling Cylinders

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