Class Notes Physics I, 11/30/98

SHM of floating objects

The Bobbing Buoy

A uniform cylindrical buoy will, when pushed downward and released, bob up and down in response to changes in the buoyant force, which we show constitutes a linear restoring force F = - k x. We determine k experimentally by adding a known weight to the buoy and observing the change in its equilibrium position, then theoretically by determining the buoyant force corresponding to an additional immersion by distance `dx. We compare experimental and theoretical results to the observed period of motion.

Floating CD Jewel Box: Angular SHM

When an object rotating about a fixed axis experiences a linear restoring net torque `tau = - k `theta, it will undergo simple harmonic motion with angular frequency 'omega `sqrt ( k / I), where I is its moment of inertia. We analyze the angular simple harmonic motion of a floating rectangular piece of Styrofoam on which known masses have been placed at known distances from the center of mass. We observe the angular displacement resulting from a known torque and use this observation to determine k. From the positions of the known masses we determine the moment of inertia about the central axis of the rectangle. We determine the resulting theoretical angular frequency and compare with the observed angular frequency of the motion.

We note that a change in the scale of the system (changing the dimensions of the Styrofoam and still placing the washers at its corners), with no change in the masses of the washers apparently results in no change in the period of motion. We analyze the scaling factors involved and raise some questions about our results.

The Bobbing Buoy

• If the dowel is immersed deeper in the water, the additional buoyant force on the newly immersed segment of the dowel will constitute a restoring force tending to push the dowel back toward its equilibrium position.
• If the dowel is raised partially out of the water, the buoyant force decreases and does not fully counter the gravitational force so that a net downward force now tends to restore the system to its equilibrium position.
• If the dowel is uniform, then the change in the buoyant force will be proportional to the distance the dowel is raised or lowered from its equilibrium position.
• The magnitude of the restoring force will therefore be proportional to the displacement from equilibrium, and we will have a linear restoring force

• The negative sign reflects the fact that the net force is in the opposite direction to the displacement of the system from its equilibrium position.
• That is, if the system is displaced downward the additional buoyant force is upward and if the system is displaced upward the reduced buoyant force results in a net downward force.

• The result of the gravitational force on this added mass is that the system sinks to a new equilibrium position 1.25 cm, or .0125 m, below the original.
• Since the gravitational force on a mass of .2 grams is .00196 N, we conclude that an additional buoyant force of .00196 N (this value rounded to one significant figure) is acting upward on the original system.
• Thus a displacement of -.0125 m results in a restoring force of .00196 N (note that by the signs we have implicitly chosen the upward direction as positive).

We therefore conclude that the constant k of the relationship F = - kx is k = .157 N / m.

• We note that our observation of the additional depth has perhaps a 10% margin of error, so that the three apparently significant figures obtained for k are not in fact all significant.

By weighing the dowel and washer we determine that its mass is 5.5 grams, or .0055 kg.

• We therefore conclude from the linearity of its restoring force that the system will tend to undergo simple harmonic motion was a new frequency `omega = `sqrt(k/m) = 5.3 rad / s.
• Its period of SHM will therefore be 2 `pi / (5.3 rad/s) = 1.2 sec, approximately.

This result is not in good agreement with the period observed when the dowel is pushed downward then released.

• We observed the period to be the time required for the dowel to rise and fall.  
• The observed period was around .7 sec.

Calculating the Buoyant Restoring Force

• When an additional segment of length `dx (shaded ingredient in the figure below) is immersed in the water, the resulting buoyant force will by Archimedes' Principle be equal to the weight of the water displaced.
• To find the weight of the water displaced we find the volume of water displaced and multiplied by the density of water.

• We displace a volume `dV equal to that of a cylinder whose diameter is 6.35 mm and whose altitude is `dx.
• This volume will be equal to the product of the area `pi r^2 of the base of the cylinder and its altitude `dx.
• Using half the diameter of the dowel as the radius, we find that `dV = A `dx = 31 mm^2 `dx = 3.1 * 10^-5 m^2 `dx.

The weight of the water displaced, and thus the restoring force will therefore have magnitude `rho g `dV, so

• | `dF | = 1000 kg / m^3 * (9.8 m/s^2) * 3.1 * 10^-5 m^2 `dx = .3 N / m * `dx.

Since the direction of the restoring forces opposite to the direction of `dx, we see that

• `dF = -3 N/m `dx.

We therefore conclude that k = -`dF / `dx = 3 N / m (or more accurately .294 N/m).

Using this theoretical value of k we obtain `omega = 9.2 rad/s, so that the predicted period is T = .67 sec.

• This period is in good agreement with the observed period.
• This should motivate us to go back and repeat our observations to determine k experimentally.

• We note that it is a bit tricky to measure the displacement resulting from an added weight on the top of the dowel and that our first attempt might have been somewhat clumsy.
• We note also that the motion of the dowel is not perfect SHM, since the frictional drag of the water rapidly decreases the amplitude of the motion.  As a result the actual frequency will differ somewhat from the 'ideal' predicted frequency, which did not take account of this drag factor.

Floating CD Jewel Box: Angular SHM

• When one end or another is displaced downward then released the ends alternatively bob up and down, with one end up when the other is down.
• This motion can be understood as rotation about an axis perpendicular to the length of the rectangle and parallel to the water surface.

• By analogy with the simple harmonic motion of a pendulum, a weight on a spring or a bobbing buoy, which are the result of the restoring force F = - kx, the present system will be governed by a restoring torque of the form `tau = -k `theta.
• That is, the system will experience a restoring torque `tau in response to an angular displacement `d`theta, with the torque and angular displacement related by `tau = - k `theta, where `theta is the angular displacement from the equilibrium position.

Calculus-based note justifying this analogy:

• The result of such a restoring torque is that `alpha = -k `theta / I.
• Since `alpha = d^2(`theta) / `dt^2, the second derivative of the angular position function `theta, we obtain the differential equation identical except for the symbols used for the variables to that we solved for the standard SHM situation.
• The details are in the University Physics text.

• The total mass added to each end of the system is 60 grams, corresponding to a set of two 15 g washers at each corner of the rectangle.
• The separation of the centers of mass of these washers is approximately 8 cm. This configuration is depicted in the upper right-hand corner of the figure below.

• The distance is observed to be 4 cm.
• The torque is easily calculated as in the figure below to be `d`tau = .006 N.

The angular displacement corresponding to a .5 cm displacement at a distance of 5 cm from the axis of rotation is `d`theta = tan^-1(.5 cm / 5 cm) = .1 rad.

• By analogy with linear SHM, we conjecture that since the moment of inertia I is analogous to mass m (recall the two forms of Newton's Second Law F = ma and `tau = I `alpha), for rotational SHM we should have `omega = `sqrt( k / I ).
• This is in fact the case (241 students note that this follows from the solution of the differential equation mentioned above).

• Ignoring the relatively small moment of inertia of the Styrofoam rectangle, we calculate the moment of inertia of just the washers.
• Making the approximation that the mass of each washer is concentrated at its center, we easily find the sum of the mr^2 contributions of the washers (each centered at a point 4 cm from the axis of rotation) to be

I = .000192 kg m^2 and k = .06 m N / rad imply `omega = 17.6 rad / s.

• You should verify this result.

• It turns out that the moment of inertia of the Styrofoam, calculated as a rod rotating about its center, is only about 2 percent that of the washers.
• Since the angular frequency was calculated using the square root of the moment of inertia, adding the moment of inertia of the Styrofoam will change our prediction by only about 1% and will not explain the discrepancy.
• A better explanation is the difficulty of accurately observing the apparent .5 cm displacement of the end of the Styrofoam rectangle. There was easily a 10% margin of error in that observation.

• to increase the distance of each washer from the center from 4 cm to 7.5 cm, which would increase the moment of inertia by a factor of (7.5 / 4)^2

• to change the angular displacement of the end resulting from the addition of the 15 g washer by factor (4 / 7.5), since the distance from the axis of rotation would change by factor 7.5 / 4
• to change the torque associated with the 15 g washer by factor of (7.5 / 4)

• change its moment of inertia by factor c^2 (recall that moment of inertia `sum (mr^2) is proportional to the square of the distance of the masses from the axis of rotation),

• change the angular displacement corresponding to a given torque by factor 1/c,
• change the torque itself by factor c, and therefore
• change k = `tau / `theta by factor c^2

Or is this really the case?

• Since there is more Styrofoam to be submerged, would the displacement of the end due to the addition of the 15 g mass also change by factor 1/c (4 / 7.5 in our example)?  And would not the width of the rectangle have an effect?
• If so what would this do to our predicted angular frequency and period?
• How could we test our alternative hypotheses?