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A uniform disc A, of mass 2.0 kg and radius 0.20 m, turns freely in a horizontal plane on a smooth…

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

A uniform disc A, of mass $2.0\ \text{kg}$ and radius $0.20\ \text{m}$ , turns freely in a horizontal plane on a smooth vertical axle at $30\ \text{rad/s}$ . A second uniform disc B, of mass $1.0\ \text{kg}$ and radius $0.10\ \text{m}$ , not turning, is gently placed on A, on the same axle, as shown in the figure. B can also turn freely on the axle, and its weight presses evenly over its whole face. The coefficient of sliding friction between the discs is $0.25$ . Take $g = 10\ \text{m/s}^2$ .

Side view of a vertical axle. Disc A, of mass 2.0 kg and radius 0.20 m, turns with the axle at omega1 = 30 rad/s. Above it, disc B, of mass 1.0 kg and radius 0.10 m, is centred on the same axle and is about to be lowered onto A.

What is the final common angular speed of the discs?
How much energy turns into heat?
Find the friction torque between the discs while they slip, and the time they take to reach the common angular speed.

The worked solution

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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.

Given. $I_A = \tfrac12 m_A R_A^2 = 0.040\ \text{kg m}^2$ , $I_B = \tfrac12 m_B R_B^2 = 0.005\ \text{kg m}^2$ .

Step 1. Common speed. The friction torques on A and B are equal and opposite, and the smooth axle gives no torque. So the total angular momentum about the axle stays the same:

\displaystyle \omega = \frac{I_A\omega_1}{I_A + I_B} = \frac{1.20\ \text{kg m}^2/\text{s}}{0.045\ \text{kg m}^2} = 26.7\ \text{rad/s}

Step 2. Heat. $\tfrac12 I_A\omega_1^2 - \tfrac12(I_A + I_B)\omega^2 = 18.0\ \text{J} - 16.0\ \text{J} = 2.0\ \text{J}$ .

Step 3. Friction torque. B's weight spreads evenly, so the pressure on its face is $p = \dfrac{m_B g}{\pi R_B^2}$ . A thin ring of radius $s$ and width $ds$ has area $2\pi s\,ds$ . Friction on it is $\mu p\,2\pi s\,ds$ , with a torque $s$ times that. Adding the rings:

\displaystyle \tau = \int_0^{R_B} \mu p\,2\pi s^2\,ds = \tfrac23\mu m_B g R_B = \tfrac23 \times 0.25 \times 1.0\ \text{kg} \times 10\ \text{m/s}^2 \times 0.10\ \text{m} = 0.167\ \text{N m}

Step 4. Time. B's angular acceleration is $\dfrac{\tau}{I_B} = 33.3\ \text{rad/s}^2$ . It reaches $26.7\ \text{rad/s}$ after $t = \dfrac{26.67\ \text{rad/s}}{33.33\ \text{rad/s}^2} = 0.80\ \text{s}$ .

Answer: (a) $26.7\ \text{rad/s}$ . (b) $2.0\ \text{J}$ . (c) $0.167\ \text{N m}$ ; $0.80\ \text{s}$ .

Check. A slows at $\dfrac{\tau}{I_A} = 4.17\ \text{rad/s}^2$ . After $0.80\ \text{s}$ it turns at $30 - 4.17 \times 0.80 = 26.7\ \text{rad/s}$ , the same as B.

Watch out. Friction acts at every radius from the centre out to $R_B$ , not only at the rim. Taking the torque as $\mu m_B g R_B$ would give a time of $0.53\ \text{s}$ , too short.

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