A useful technique for calculating rotational inertia about rotational axes for shapes whose rotational inertia about an axis through their center of mass is known is the parallel-axis theorem.
where Icm is the rotational inertia about an axis through the center of mass of the rigid body, h is the perpendicular distance (this is also the distance of closest approach) from the center of mass of an axis parallel to the axis for which Icm is known, and I is the rotational inertia about that parallel axis.
The proof of this theorem is in the Interactive Textbook so we will not reproduce it here. We should do an example of how it can be used however.
Find the rotational inertia of the barbell arrangement shown below in which to solid spheres of mass M and radius R are attached by a rigid, thin rod of mass m (m = ½ M) and length b (b = 2R). The barbell is rotating around an axis through the center of one of the spheres as shown in the figure below.
Solution: We start by using the parallel-axis theorem to get the rotational inertia for the spheres about the axis. For the sphere through which the rotation axis penetrates (call this mass 1), we have
For mass 2, the sphere on the right, we use the parallel-axis theorem to get
For the rod, mass 3, the rotational inertia is
The combined rotational inertia is the sum of all three contributions, so
where q is the angle between the vector r from the rotation axis to the point P where force F acts.
Note that the angle q is defined as the angle from r to F. To get the direction of the torque, we follow the convention of the right-hand rule (to be consistent with our definition of directions for a and w) in letting the curl of fingers on the right hand in the direction of r to F and setting the direction of t as the direction our thumb on the right-hand points. In math-speak this is all condensed down to
The terminology is that the torque is equal to the cross product of vector r and F. To get a great visual cue as to what the cross product does to produce a third vector from two other vectors, check out the great A 3-D cross product visualization applet from the Syracuse University physics department. The second of these links takes you to a page of simulations on kinematics, gravity, and other topics along with 3-D visualization of other physics concepts. It's lots of fun! If we look at a single particle undergoing circular motion but under the influence of a non-zero tangential force as in the figure below, then we can also relate the torque to angular acceleration. In this case there must be both a radial force component (pointed inward so actually antiparallel to the radius) to maintain the circular motion and the tangential force component we assumed. The tangential force component, Ft, leads to a tangential acceleration of the mass. We can relate that to the angular acceleration since at = ra By our definition of torque, we must have t = Ftr = (mat)r = mr2a The mass times radius squared term should look familiar. It is just the rotational inertia for a point mass rotating around an axis a distance r away! Hence we generalize our definition by considering any solid object as made of a large number of infinitesimal mass elements, all undergoing the same angular acceleration about a fixed axis due to the application of a force to the object. Then, å t = óõ (r2 dm)a = a óõ r2 dm = Ia As usual, we should follow the derivation with an example. Example 3: A pulley consists of a uniform thin disk with mass M and radius R which is fastened to a desk so that the axle through the center of the disk is allowed to turn freely and without friction. An ideal string is wrapped around the disk then attached to a block of mass m (m = ½ M). Initially, the block is held, but is released at time t = 0. Find the acceleration of the block and the tension in the string. Solution: As you can see, we have stripped back our half-Atwood machine to just one block and the pulley now to emphasize the use of our rotational dynamical equations. In this case, the force diagram appears as shown below. The equations of motion for the block and the disk are block: mg - T = ma disk: t = R*T = Ia = ½ MR2(a/R) Note that in the above we assumed something, namely that the string must unwrap from the disk at the same rate as the block is falling to keep the string from stretching and that the angular of the wheel is connected mathematically to that angular rate. This only works if the string is truly ideal, i.e. doesn't take up any space while wrapped around the edge of the disk. We can solve the two equations above for the two unknowns to get R*T = ½ MR2(a/R) T = ½ Ma From the equation of motion for the block, we have T = ½ Mg - ½ Ma so we get ½ Ma = ½ Mg - ½ Ma ==> a = ½ g The tension in the string is therefore T = ½ Ma = ¼ Mg Connect Question! What would the tension in the string for the previous problem have been if the disk of the pulley were massless? Connect Question! The end of a meter stick is placed into a pivot on the floor so that it can rotate freely (without friction) about that end. You hold the meter stick vertically initially, then release it with just a tiny push so that it begins to fall. What is the angular acceleration of the meter stick when the stick makes an angle q with respect to the vertical? We can get the solution to the above by noting that only the gravitational force can produce a torque. Normal forces acting at the pivot point cannot because the axis of rotation goes through that point and hence the lever arm, or r vector has zero magnitude for all such forces. For gravity acting at the center of mass (see force diagram below), we have the torque as Mg(b/2) sinq. The rotational inertia of a thin cylinder about an end is (1/3) Mb2, so T = Ia ½ Mgb sinq = (1/3)Mb2a ==> a = (3/2)(g/b) sinq Statics One last topic we can address using the tools we have developed thus far is static equilibrium. This actually covers much of the world we generally view since most things are arranged so that they are static: books on bookshelves, signs hanging from buildings, buildings, trees, etc. all have to have their structures arranged so that they maintain a constant orientation despite external forces acting. The actual conditions necessary to maintain static equilibrium are The net external forces acting must equal zero. The net external torque about any axis must be zero. The first condition guarantees translational equilibrium. The linear acceleration of the system must be zero when observed from any inertial frame. The second gives us rotational equilibrium. The system is not rotating with respect to any axis in an inertial frame. So there are no new concepts here, just more experience in using the rotational and translational equations of motion. Example 4: A person holds a 50.0 Nt weight in her hand. The forearm is being held horizontal. The bicep muscle is attached 3.00 cm from the joint and the weight is 35.0 cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of the forearm. Solution: Let's look at the forces from a simpler perspective of just two massless bars for the forearm and upper arm. Then the forces acting are as shown. Let up be the positive direction and counterclockwise rotation be the positive rotation direction. Then, for static equilibrium, we want forces: F - R - mg = 0 torques: Fd - mgb = 0 where we have chosen the elbow as the pivot point (the rotation axis goes through there) for calculating the torque. We can plug in numbers to the torque equation to find that F(0.03 m) - (50.0 Nt)(0.35 m) = 0 ==> F = 583 Nt Plugging this back into our force equation gives R = 533 Nt. These magnitudes correspond to 131 pounds and 120 pounds for F and R, respectively. The forces at joints and in muscles of the body can be very large. If we actually use the fact that the bicep makes an angle of 15.0° with respect to the vertical, then F has both vertical and horizontal components. In this case the force equations would be y: Fy - Ry - mg = 0 x: -Fx + Rx = 0 The horizontal components for F and R clearly must be equal. The torque equation is unchanged since the net force in the x direction is zero. Then Fyd = mgb ==> Fy = 583 Nt Since we know that Fy = F cos15° ==> F = 604 Nt and Fx = F sin15° = 156 Nt = Rx. As before, we have Ry = 533 Nt so the total force on the elbow is sqrt[Rx2 + Ry2] = 555 Nt.
If we look at a single particle undergoing circular motion but under the influence of a non-zero tangential force as in the figure below, then we can also relate the torque to angular acceleration.
In this case there must be both a radial force component (pointed inward so actually antiparallel to the radius) to maintain the circular motion and the tangential force component we assumed. The tangential force component, Ft, leads to a tangential acceleration of the mass. We can relate that to the angular acceleration since
By our definition of torque, we must have
The mass times radius squared term should look familiar. It is just the rotational inertia for a point mass rotating around an axis a distance r away! Hence we generalize our definition by considering any solid object as made of a large number of infinitesimal mass elements, all undergoing the same angular acceleration about a fixed axis due to the application of a force to the object. Then,
As usual, we should follow the derivation with an example.
A pulley consists of a uniform thin disk with mass M and radius R which is fastened to a desk so that the axle through the center of the disk is allowed to turn freely and without friction. An ideal string is wrapped around the disk then attached to a block of mass m (m = ½ M). Initially, the block is held, but is released at time t = 0. Find the acceleration of the block and the tension in the string.
Solution: As you can see, we have stripped back our half-Atwood machine to just one block and the pulley now to emphasize the use of our rotational dynamical equations. In this case, the force diagram appears as shown below.
The equations of motion for the block and the disk are
Note that in the above we assumed something, namely that the string must unwrap from the disk at the same rate as the block is falling to keep the string from stretching and that the angular of the wheel is connected mathematically to that angular rate. This only works if the string is truly ideal, i.e. doesn't take up any space while wrapped around the edge of the disk.
We can solve the two equations above for the two unknowns to get
From the equation of motion for the block, we have
so we get
The tension in the string is therefore
We can get the solution to the above by noting that only the gravitational force can produce a torque. Normal forces acting at the pivot point cannot because the axis of rotation goes through that point and hence the lever arm, or r vector has zero magnitude for all such forces. For gravity acting at the center of mass (see force diagram below), we have the torque as Mg(b/2) sinq.
The rotational inertia of a thin cylinder about an end is (1/3) Mb2, so
One last topic we can address using the tools we have developed thus far is static equilibrium. This actually covers much of the world we generally view since most things are arranged so that they are static: books on bookshelves, signs hanging from buildings, buildings, trees, etc. all have to have their structures arranged so that they maintain a constant orientation despite external forces acting. The actual conditions necessary to maintain static equilibrium are
The first condition guarantees translational equilibrium. The linear acceleration of the system must be zero when observed from any inertial frame. The second gives us rotational equilibrium. The system is not rotating with respect to any axis in an inertial frame. So there are no new concepts here, just more experience in using the rotational and translational equations of motion.
A person holds a 50.0 Nt weight in her hand. The forearm is being held horizontal. The bicep muscle is attached 3.00 cm from the joint and the weight is 35.0 cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of the forearm.
Solution: Let's look at the forces from a simpler perspective of just two massless bars for the forearm and upper arm. Then the forces acting are as shown.
Let up be the positive direction and counterclockwise rotation be the positive rotation direction. Then, for static equilibrium, we want
where we have chosen the elbow as the pivot point (the rotation axis goes through there) for calculating the torque. We can plug in numbers to the torque equation to find that
Plugging this back into our force equation gives R = 533 Nt. These magnitudes correspond to 131 pounds and 120 pounds for F and R, respectively. The forces at joints and in muscles of the body can be very large.
If we actually use the fact that the bicep makes an angle of 15.0° with respect to the vertical, then F has both vertical and horizontal components.
In this case the force equations would be
The horizontal components for F and R clearly must be equal. The torque equation is unchanged since the net force in the x direction is zero. Then
Since we know that Fy = F cos15° ==> F = 604 Nt and Fx = F sin15° = 156 Nt = Rx. As before, we have Ry = 533 Nt so the total force on the elbow is sqrt[Rx2 + Ry2] = 555 Nt.