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Newton's Second Law for Rotation

Acceleration produced when a force acts on a mass (F = ma) can be applied to rotating systems as well.

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Newton's Second Law for Rotation

Students will learn to apply Newton's 2nd law in the case of rotational dynamics.

Key Equations

\begin{align*}\alpha = \tau_{net} / I\end{align*}

Angular accelerations are produced by net torques,with inertia opposing acceleration; because \begin{align*}\alpha \end{align*} is the rotational version of a, linear acceleration, \begin{align*} \tau \end{align*} is the rotational version of F, force, and I is the rotational version of mass.

This is the rotational analog of Newton's 2nd law \begin{align*}a = F_{net} / m\end{align*}

\begin{align*}\tau_{net} = \Sigma \tau_i = I \alpha\end{align*}

The net torque is the vector sum of all the torques acting on the object. When adding torques it is necessary to subtract CW from CCW torques.

  • Use this law just as you did in the Newton's Laws lessons. First choose a pivot point to take all torques around, and then add up all the torques acting on an object and that will equal the moment of inertia multiplied by the angular acceleration.
  • Torques produce angular accelerations, but just as masses resist acceleration (due to inertia), there is an inertia that opposes angular acceleration. The measure of this inertial resistance depends on the mass, but more importantly on the distribution of the mass in a given object. The moment of inertia, \begin{align*}I,\end{align*} is the rotational version of mass. Values for the moment of inertia of common objects are given above. Torques have only two directions: those that produce clockwise (CW) and those that produce counterclockwise (CCW) rotations. The angular acceleration or change in \begin{align*}\omega\end{align*} would be in the direction of the torque.
  • Many separate torques can be applied to an object. The angular acceleration produced is \begin{align*}\alpha = \tau_{net}/I\end{align*}

Example 1

Example 2

Some old doors have the door knob in the center of the door like in the picture below. If you had a door of mass 20 kg and 1.5 m wide, what would be the angular acceleration if you pushed at the center of the door with a force of 50 N? What would the angular acceleration be if you pushed at the far end of the door with the same force? Where should you push the door if you want to open it the fastest?


To solve this problem, we will plug in the known values into Newton's second law for rotation and solve for \begin{align*}\alpha\end{align*}. When calculating the moment of inertia of the door, we'll consider it a rod being rotated about one end. We'll first calculate the angular acceleration when pushing from the center of the door.

\begin{align*} \Sigma\tau&=I\alpha\\ Fr&=I\alpha\\ Fr&=\frac{1}{3}mr^2\alpha\\ \alpha&=\frac{3Fr}{mr^2}\\ \alpha&=\frac{3*50\;\text{N}*.75\;\text{m}}{20\;\text{kg}*(1.5\;\text{m})^2}\\ \alpha&=2.5\;\text{rad/s}^2\\ \end{align*}

Now we'll calculate the angular acceleration when pushing from the end of the door.

\begin{align*} \alpha&=\frac{3Fr}{mr^2}\\ \alpha&=\frac{3*50\;\text{N}*1.5\text{m}}{20\;\text{kg}*(1.5\;\text{m})^2}\\ \alpha&=5\;\text{rad/s}^2\\ \end{align*}

Clearly, it is much faster and easier to open doors when pushing from the point farthest from the hinge.

Watch this Explanation

Time for Practice

  1. In the figure we have a horizontal beam of length, \begin{align*}L\end{align*}, pivoted on one end and supporting \begin{align*}2000 \;\mathrm{N}\end{align*} on the other. Find the tension in the supporting cable, which is at the same point at the weight and is at an angle of \begin{align*}30\end{align*} degrees to the vertical. Ignore the weight of the beam.
  2. Two painters are on the fourth floor of a Victorian house on a scaffold, which weighs \begin{align*}400 \;\mathrm{N}\end{align*}. The scaffold is \begin{align*}3.00 \;\mathrm{m}\end{align*} long, supported by two ropes, each located \begin{align*}0.20 \;\mathrm{m}\end{align*} from the end of the scaffold. The first painter of mass \begin{align*}75\;\mathrm{kg}\end{align*} is standing at the center; the second of mass, \begin{align*}65.0 \;\mathrm{kg}\end{align*}, is standing \begin{align*}1.00 \;\mathrm{m}\end{align*} from one end.
    1. Draw a free body diagram, showing all forces and all torques. (Pick one of the ropes as a pivot point.)
    2. Calculate the tension in the two ropes.
    3. Calculate the moment of inertia for rotation around the pivot point, which is supported by the rope with the least tension. (This will be a compound moment of inertia made of three components.)
    4. Calculate the instantaneous angular acceleration assuming the rope of greatest tension breaks.
  3. A horizontal beam of weight \begin{align*}60 \;\mathrm{N}\end{align*} and \begin{align*}1.4 \;\mathrm{m}\end{align*} in length has a \begin{align*}100 \;\mathrm{N}\end{align*} weight on the end. It is supported by a cable, which is connected to the horizontal beam at an angle of \begin{align*}37\end{align*} degrees at \begin{align*}1.0 \;\mathrm{m}\end{align*}from the wall. Further support is provided by the wall hinge, which exerts a force of unknown direction, but which has a vertical (friction) component and a horizontal (normal) component.
    1. Find the tension in the cable.
    2. Find the two components of the force on the hinge (magnitude and direction).
    3. Find the coefficient of friction of wall and hinge.

  4. On a busy intersection a \begin{align*}3.00 \;\mathrm{m}\end{align*} beam of \begin{align*}150 \;\mathrm{N}\end{align*} is connected to a post at an angle upwards of \begin{align*}20.0\end{align*} degrees to the horizontal. From the beam straight down hang a \begin{align*}200 N\end{align*} sign \begin{align*}1.00 \;\mathrm{m}\end{align*} from the post and a \begin{align*}500\;\mathrm{N}\end{align*} signal light at the end of the beam. The beam is supported by a cable, which connects to the beam \begin{align*}2.00 \;\mathrm{m}\end{align*} from the post at an angle of \begin{align*}45.0\end{align*}degrees measured from the beam; also by the hinge to the post, which has horizontal and vertical components of unknown direction.
    1. Find the tension in the cable.
    2. Find the magnitude and direction of the horizontal and vertical forces on the hinge.
    3. Find the total moment of inertia around the hinge as the axis.
    4. Find the instantaneous angular acceleration of the beam if the cable were to break.

  5. The medieval catapult consists of a \begin{align*}200 \;\mathrm{kg}\end{align*} beam with a heavy ballast at one end and a projectile of \begin{align*}75.0 \;\mathrm{kg}\end{align*} at the other end. The pivot is located \begin{align*}0.5 \;\mathrm{m}\end{align*} from the ballast and a force with a downward component of \begin{align*}550 \;\mathrm{N}\end{align*} is applied by prisoners to keep it steady until the commander gives the word to release it. The beam is \begin{align*}4.00 \;\mathrm{m}\end{align*} long and the force is applied \begin{align*}0.900 \;\mathrm{m}\end{align*}from the projectile end. Consider the situation when the beam is perfectly horizontal.
    1. Draw a free-body diagram labeling all torques.
    2. Find the mass of the ballast.
    3. Find the force on the vertical support.


  1. \begin{align*}2300\;\mathrm{N}\end{align*}
  2. b. \begin{align*}771\;\mathrm{N}, 1030\;\mathrm{N}\end{align*} c. \begin{align*}554 \;\mathrm{kgm}^2\end{align*} d. \begin{align*}4.81\mathrm{rad/sec}^2\end{align*}
  3. a. \begin{align*}300\;\mathrm{N}\end{align*} b. \begin{align*}240N, -22\;\mathrm{N}\end{align*} c. \begin{align*}.092\end{align*}
  4. a. \begin{align*}2280\;\mathrm{N}\end{align*} b. \begin{align*}856 \;\mathrm{n}\end{align*} toward beam, \begin{align*}106\;\mathrm{N}\end{align*} down c. \begin{align*}480 \;\mathrm{kgm}^2\end{align*} d. \begin{align*}3.8 \;\mathrm{rad/sec}^2\end{align*}
  5. b. \begin{align*}345 \;\mathrm{kg} \end{align*} using \begin{align*} 9.8 \;\mathrm{m/s}^2\end{align*} for acceleration of gravity c. 4690 N

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