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A solid cylinder of mass 5.0 kg and radius 0.15 m stands on a rough horizontal floor…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solid cylinder of mass $5.0\ \text{kg}$ and radius $0.15\ \text{m}$ stands on a rough horizontal floor. Its axle is joined through a smooth bearing to a light horizontal spring of force constant $240\ \text{N/m}$ , whose other end is fixed to a wall, as in the figure. The cylinder is rolled aside and released from rest. The coefficient of static friction between the cylinder and the floor is $0.25$ . Take $g = 10\ \text{m/s}^2$ .

A solid cylinder of mass 5.0 kg and radius 0.15 m stands on a rough floor, with a coefficient of static friction of 0.25. A horizontal spring of force constant 240 N/m joins a smooth bearing on the cylinder's axle to a wall on the left. A curved arrow shows that the cylinder can roll.

While the cylinder rolls without slipping, what is the period of its oscillation?
What is the largest amplitude for which it never slips?
At that amplitude, what is the greatest speed of the axle?
At that amplitude, what is the greatest kinetic energy of rotation of the cylinder?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the effective mass from the kinetic energy, then the friction the rolling needs, and set its largest value equal to $\mu Mg$ .

Step 1. Period (a). The kinetic energy is $\tfrac12 Mv^2 + \tfrac12\left(\tfrac12 MR^2\right)\left(\tfrac{v}{R}\right)^2 = \tfrac12\left(\tfrac32 M\right)v^2$ . So the cylinder oscillates like a mass of $\tfrac32 M = 7.5\ \text{kg}$ on the spring:

\displaystyle \omega^2 = \frac{2k}{3M} = 32\ \text{s}^{-2}, \quad T = \frac{2\pi}{\omega} = 1.111\ \text{s}

Step 2. The friction needed. With $f$ the friction on the cylinder: $Ma = -kx - f$ for the axle and $\tfrac12 MR^2 \cdot \dfrac{a}{R} = fR$ for the spin, so $f = \tfrac12 Ma$ . Together these give $f = -\tfrac13 kx$ , largest at $x = A$ .

Step 3. Largest amplitude (b). $\tfrac13 kA \le \mu Mg$ gives

\displaystyle A_{\max} = \frac{3\mu Mg}{k} = \frac{3 \times 0.25 \times 5.0\ \text{kg} \times 10\ \text{m/s}^2}{240\ \text{N/m}} = 15.6\ \text{cm}

Step 4. Greatest speed (c). $v_{\max} = \omega A_{\max} = 5.657\ \text{rad/s} \times 0.1563\ \text{m} = 0.884\ \text{m/s}$ .

Step 5. Kinetic energy of rotation (d). It is $\tfrac12\left(\tfrac12 MR^2\right)(v/R)^2 = \tfrac14 Mv^2$ , largest at the centre: $\tfrac14 \times 5.0\ \text{kg} \times (0.884\ \text{m/s})^2 = 0.977\ \text{J}$ .

Answer: (a) $1.111\ \text{s}$ . (b) $15.6\ \text{cm}$ . (c) $0.884\ \text{m/s}$ . (d) $0.977\ \text{J}$ .

Check. The energy stored at the ends is $\tfrac12 kA^2 = 2.930\ \text{J}$ . At the centre it is all kinetic, and one third of it is rotation, $0.977\ \text{J}$ , as for any solid cylinder rolling. The radius drops out of (a) to (d), though the spin itself, $5.89\ \text{rad/s}$ at the centre, does depend on it.

Watch out. The friction does no work here, since the point of contact does not slide, but it is not zero. Leaving it out of the axle's equation gives the period of a block, $2\pi\sqrt{M/k}$ .

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