Angular Momentum and Its Conservation – IP

Example 10.13 Calculating the Torque Putting Angular Momentum Into a Lazy Susan

Figure 10.21 shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan's 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk?

Figure 10.21 A partygoer exerts a torque on a lazy Susan to make it rotate. The equation $\text { net } \tau=\frac{\Delta L}{\Delta t}$ gives the relationship between torque and the angular momentum produced.

Strategy

We can find the angular momentum by solving $\tau=\frac{\Delta L}{\Delta t} \text { for } \Delta L$ , and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, $\Delta L=L$. To find the final velocity, we must calculate ω from the definition of $L$ in $L=I \omega$.

Solution for (a)

Solving $\text { net } \tau=\frac{\Delta L}{\Delta t} \text { for } \Delta L$ gives

$\Delta L=(\operatorname{net} \tau) \Delta \mathrm{t}$

Because the force is perpendicular to $r$, we see that $\text { net } \tau=r F \text {, }$, so that

$\begin{aligned} L &=\mathrm{rF} \Delta t=(0.260 \mathrm{~m})(2.50 \mathrm{~N})(0.150 \mathrm{~s}) \\ &=9.75 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} \end{aligned}$

Solution for (b)

The final angular velocity can be calculated from the definition of angular momentum,

$L=I \omega$

Solving for $\omega$ and substituting the formula for the moment of inertia of a disk into the resulting equation gives

$\omega=\frac{L}{I}=\frac{L}{\frac{1}{2} M R^{2}}$

And substituting known values into the preceding equation yields

$\omega=\frac{9.75 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}}{(0.500)(4.00 \mathrm{~kg})(0.260 \mathrm{~m})^{2}}=0.721 \mathrm{rad} / \mathrm{s}$

Discussion

Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan.