10.7 Newton’s second law for rotation  (Page 2/5)

Deriving newton’s second law for rotation in vector form

As before, when we found the angular acceleration, we may also find the torque vector. The second law $\Sigma \overrightarrow{F}=m\overrightarrow{a}$ tells us the relationship between net force and how to change the translational motion of an object. We have a vector rotational equivalent of this equation, which can be found by using [link] and [link] . [link] relates the angular acceleration to the position and tangential acceleration vectors:

We form the cross product of this equation with $\overrightarrow{r}$ and use a cross product identity (note that $\overrightarrow{r}·\overrightarrow{\alpha }=0$ ):

We now form the cross product of Newton’s second law with the position vector $\overrightarrow{r},$

Identifying the first term on the left as the sum of the torques, and $m{r}^2$ as the moment of inertia, we arrive at Newton’s second law of rotation in vector form:

This equation is exactly [link] but with the torque and angular acceleration as vectors. An important point is that the torque vector is in the same direction as the angular acceleration.

Applying the rotational dynamics equation

Before we apply the rotational dynamics equation to some everyday situations, let’s review a general problem-solving strategy for use with this category of problems.

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

Strategy

Solution

1. The moment of inertia of a solid disk about this axis is given in [link] to be
   
   $\frac{1}{2}M{R}^2.$
   
   We have $M=50.0\text{kg}$ and $R=1.50\text{m}$ , so
   
   $I=(0.500)(50.0\text{kg}){(1.50\text{m})}^2=56.25{\text{kg-m}}^2.$
   
   To find the net torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that
   
   $\tau =rF\text{sin}\theta =(1.50\text{m})(250.0\text{N})=375.0\text{N-m}.$
   
   Now, after we substitute the known values, we find the angular acceleration to be
   
   $\alpha =\frac{\tau }{I}=\frac{375.0\text{N-m}}{56.25{\text{kg-m}}^2}=6.67\frac{\text{rad}}{{\text{s}}^2}.$
2. 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 approximating the child as a point mass at a distance of 1.25 m from the axis. Then
   
   $I_{\text{c}}=m{R}^2=(18.0\text{kg}){(1.25\text{m})}^2=28.13{\text{kg-m}}^2.$
   
   The total moment of inertia is the sum of the moments of inertia of the merry-go-round and the child (about the same axis):
   
   $I=28.13{\text{kg-m}}^2+56.25{\text{kg-m}}^2=84.38{\text{kg-m}}^2.$
   
   Substituting known values into the equation for α gives
   
   $\alpha =\frac{\tau }{I}=\frac{375.0\text{N-m}}{84.38{\text{kg-m}}^2}=4.44\frac{\text{rad}}{{\text{s}}^2}.$