<< Chapter < Page Chapter >> Page >

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 Σ F = m 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:

a = α × r .

We form the cross product of this equation with r and use a cross product identity (note that r · α = 0 ):

r × a = r × ( α × r ) = α ( r · r ) r ( r · α ) = α ( r · r ) = α r 2 .

We now form the cross product of Newton’s second law with the position vector r ,

Σ ( r × F ) = r × ( m a ) = m r × a = m r 2 α .

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:

Σ τ = I α .

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.

Problem-solving strategy: 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. Identify the pivot point. If the object is in equilibrium, it must be in equilibrium for all possible pivot points––chose the one that simplifies your work the most.
  5. Apply i τ i = I α , 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.
  6. As always, check the solution to see if it is reasonable.

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 200.0-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 friction.

Figure shows a man that pushes a merry-go-round at its edge and perpendicular to its radius.
A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque.

Strategy

The net torque is given directly by the expression i τ i = I α , To solve for α , we must first calculate the net torque τ (which is the same in both cases) and moment of inertia I (which is greater in the second case).

Solution

  1. The moment of inertia of a solid disk about this axis is given in [link] to be
    1 2 M R 2 .

    We have M = 50.0 kg and R = 1.50 m , so
    I = ( 0.500 ) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 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
    τ = r F sin θ = ( 1.50 m ) ( 250.0 N ) = 375.0 N-m .

    Now, after we substitute the known values, we find the angular acceleration to be
    α = τ I = 375.0 N-m 56.25 kg-m 2 = 6.67 rad 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 c by approximating the child as a point mass at a distance of 1.25 m from the axis. Then
    I c = m R 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 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 kg-m 2 + 56.25 kg-m 2 = 84.38 kg-m 2 .

    Substituting known values into the equation for α gives
    α = τ I = 375.0 N-m 84.38 kg-m 2 = 4 .44 rad s 2 .
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask