# 8.5 Dynamics of rotational motion: rotational inertia  (Page 3/8)

 Page 3 / 8

## Problem-solving strategy for rotational dynamics

1. Examine the situation to determine that torque and mass are involved in the rotation . Draw a careful sketch of the situation.
2. Determine the system of interest .
3. Draw a free body diagram . That is, draw and label all external forces acting on the system of interest.
4. Apply , the rotational equivalent of Newton’s second law, to solve the problem . Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
5. As always, check the solution to see if it is reasonable .

## Making connections

In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton’s second law of motion for rotation.

## Calculating the effect of mass distribution on a merry-go-round

Consider the father pushing a playground merry-go-round in [link] . He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

Strategy

Angular acceleration is given directly by the expression $\alpha =\frac{\text{net τ}}{I}$ :

$\alpha =\frac{\tau }{I}.$

To solve for $\alpha$ , we must first calculate the torque $\tau$ (which is the same in both cases) and moment of inertia $I$ (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that

$\mathrm{\tau }=\mathrm{rF}\phantom{\rule{0.25em}{0ex}}\text{sin θ}=\left(\text{1.50 m}\right)\left(\text{250 N}\right)=\text{375 N}\cdot \text{m.}$

Solution for (a)

The moment of inertia of a solid disk about this axis is given in [link] to be

$\frac{1}{2}{\text{MR}}^{2}\text{,}$

where $M=\text{50.0 kg}$ and $R=\text{1.50 m}$ , so that

$I=\left(0\text{.}500\right)\left(\text{50.0 kg}\right)\left(\text{1.50 m}{\right)}^{2}=\text{56.25 kg}\cdot {\text{m}}^{2}\text{.}$

Now, after we substitute the known values, we find the angular acceleration to be

$\alpha =\frac{\tau }{I}=\frac{\text{375 N}\phantom{\rule{0.25em}{0ex}}\text{⋅}\phantom{\rule{0.25em}{0ex}}\text{m}}{\text{56.25 kg}\cdot {\text{m}}^{2}}=\text{6.67}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}$

Solution for (b)

We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia $I$ , we first find the child’s moment of inertia ${I}_{\text{c}}$ by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

${I}_{\text{c}}={\mathrm{MR}}^{2}=\left(\text{18.0 kg}\right)\left(\text{1.25 m}{\right)}^{2}=\text{28.13 kg}\cdot {\text{m}}^{2}\text{.}$

The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of $I$ :

$I=\text{28.13 kg}\cdot {\text{m}}^{2}+\text{56.25 kg}\cdot {\text{m}}^{2}=\text{84.38 kg}\cdot {\text{m}}^{2}.$

Substituting known values into the equation for $\alpha$ gives

$\alpha =\frac{\text{τ}}{I}=\frac{\text{375 N}\cdot \text{m}}{\text{84.38 kg}\cdot {\text{m}}^{2}}=\text{4.44}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}$

Discussion

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.

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Source:  OpenStax, Introduction to applied math and physics. OpenStax CNX. Oct 04, 2012 Download for free at http://cnx.org/content/col11426/1.3
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