The Basics of Rotational Dynamics (3): Torque, Angular Acceleration & Angular Momentum

 A simple first approach to torque and its role in rotational dynamics is the topic of this post. Generally, torque is avoided in most college general physics, except for calculus-based courses. Here, I show that a regular algebra approach is feasible, as I will show in Part 4 that such an approach is possible with orbital motion. 

The most physically intuitive way to grasp torque is perhaps via the everyday example of using a wrench to turn a nut, e.g.  

Here, the simple torque equation is of relevance: 

t  =  Fr

Where is the torque, F the tangential force applies, and r the distance or radius of the force application. The red lines shown indicate the relative magnitude for the tangential force to get the wrench to turn the nut. Thus we see F2 > F1 so that much more force is needed to turn it at the much smaller radius r2. Much less force is needed at the greater distance r2.  

In many respects this is analogous to the lever principle discussed in a May, 2011 post (Fig. 5) e.g.

Here, we note the distance a needed to lift the load L is much less than for distance b or in terms of the 'law of the lever':  

F x b = L x a or F = (a/b) L 

 Thus, let L = 60 N, a = 0.5m and b = 1.5 m, then: 

F = (0.5/1.5) 60 N = 60N/3 = 20 N

But if the distances a, b had been reversed from those shown, i.e. force F applied from longer lever arm b, then we'd have found:

F = (1.5/0.5) 60 N = (3) 60N = 180 N, or three times more force needed to lift L

So no surprise the operative term is the 'moment arm' for the distance the tangential force acts from the axis of rotation.

Even without doing the math, most of those who've ever used a wrench know that the nut is more likely to turn if you apply the force F as far from the nut as possible, i.e. at position r1. Hence, we can conclude that the tendency for a force to cause a rotation increases with the distance r from the axis of rotation to the force. Given this it is useful to define the torque in the simplest sense by the equation given above.

Example Problem:

To open the door, shown in the diagram below, a tangential force F is applied at a distance r from the axis of rotation.

 If the minimum torque needed to open the door is 3.1 N· m what force must be applied if: a) r1 = 0.94m, b) r2 = 0.35m?

a) Setting t  =  F1 r1 = 3.1 N· m  we find:

F1 =  t  / r1 = 3.1 N· m/  0.94 m  =  3.3 N

b) Setting t  =  F2 r2 = 3.1 N· m  we find:

F2 =  t  / r2 = 3.1 N· m/  0.35 m  =  8.8 N

The more general case and equation would be written: 

t  =   r F (sin q)

I.e. if the force is applied at an angle q  relative to the radial line as shown in the diagram below:

In this case the force vector F needs to be resolved into radial and tangential components. Then the radial component has a magnitude of  F = r (cos q) and the tangential component has a magnitude of: F = r (sin q). But note the tangential component alone causes rotation and hence is written in terms of the torque, t :  t =  r F (sin q).  More generally, then, the torque can be defined in terms of the cross product between the vectors r and F, so:   t =  r x  F

The equivalent way to define the torque is in terms of the moment arm, denoted in the diagram as  r⊥ =  r (sin q) (From diagram)

Another problem is illustrated in the figure below. Here a roller of mass M is pulled by a rope  (with force F) wrapped around it:

A force diagram (b) is shown to the right, where  F  denotes the friction force acting opposite the rolling motion - assumed to be as shown.  The equation for the rate of change of angular momentum can be written:

RF  +  RF  =  ½ Mr ² a   where a is the angular acceleration.  Meanwhile the Newtonian (2nd law) equation is:

F  -   F    = MA 

Dividing the top equation by R and adding it to the 2nd law equation we get:

2F = 3/2 MA 

From which the acceleration:  A =  4/3 (F/M)

This is a bit surprising given it shows the center of mass acceleration is somewhat greater than it would be if the roller accelerated without any rotational motion.  Meanwhile, we have for the frictional force:

F    = - 1/3 F

Thus, since F is positive, the force of friction is actually in the backward direction rather than the forward (as assumed).  But this is just what occurs when the rear wheels of an auto accelerates.

Suggested Problems:

1) (a) What unbalanced torque will produce an angular acceleration of 3 rad/sec²  in a 100 kg disk having a radius of gyration of 0.7 m?

(b) Determine the mean effective torque that must be applied to a 40 kg flywheel with radius of gyration 0.2 m to give it an angular speed of 300 rpm in 10 secs?

2) Find the angular momentum of a 0.13 kg disk of radius 7.5 cm spinning with an angular speed of 1.15 rad/s.

3) For the example problem of the roller with friction force acting, assume a mass M = 1.5 kg and R = 0.15m.  If the angular acceleration is 1.0 rad/sec²   find:

a) The force F pulling the roller if the friction force = 0.

b) The magnitude of the force  F  if the pulling force F = 15 N.

c) The linear acceleration for case (a),  for case (b).

4) A Yo-yo is made from a uniform disk of total mass M and radius R with a string wrapped around it as shown below:

The top end of the string is tied to the ceiling and the Yo-yo is released from rest with the string vertical.

a) Draw a force diagram showing the forces acting.

b) Write the equations for the translational and rotational motions.

c) Write equations for the angular acceleration and angular momentum.

d) Find the velocity of the Yo-yo after it has dropped a height h.

5)  A cylinder of radius R = 0.5 m has a moment of inertia: I = 2.0  kg · m² about an axis of rotation. A rope is tied around the perimeter of the cylinder and a constant tension of 10 N is maintained. 

a) Find the magnitude of the torque due to the 10 N force about the axis of rotation.

b) Find the change in angular momentum of the cylinder in a time interval of 4 sec.

c) If the cylinder starts from rest what is the angular speed after 4 seconds. 

6) A torque 0f  5 N· m   is used to turn a nut with a force F = 15 N.  Find the distance from the location of applied force to the nut.

7) A light rope wrapped around a disk -shaped pulley is pulled tangentially with a force of 0.53 N. Find the torque applied to the disk and the angular acceleration of the pulley given that its mass is 1.3 kg and its radius is 0.11m.

8) In 2006 the Earth took 0.840 s  more time to complete 365 revolutions than it did in the year 1906. Find the average angular acceleration of the Earth in this time.