A Variable-Mass Snowball Rolling Down a Snowy Slope

letters to the editor

However, it is possible to find v and a analytically as a function of a different independent variable, namely the snowball’s radius r. This result is arguably even more useful than know- ing how v and a vary with time, because we can directly relate r to the distance s the snowball has rolled down the slope as

(1) using dr / d = k / 2π where is the angle through which the ball has rolled and r0 is the snowball’s initial radius. Find v (r) as follows. Solve the right-hand part of Rubin’s Eq. (12) for v and substitute it into his Eq. (13) to obtain C – 60V2 = 14rA (2) after simplifying, where V  dr/dt and A  dV/dt , and where C = 5 -1gk sin is a constant. Now observe that

(3) which is the same trick one uses to prove the work-energy theorem. Substitution of Eq. (3) into (2) results in an equation whose remaining variables r and V can be separated into

(4)

where the snowball initially has speed v = 0 and hence V = dr/dt must also be initially zero. After integrating, this result becomes

, (5)

yielding

(6)

Analytic solution for a variable-mass Finally, a can be computed as snowball (7) In the March 2019 issue, Scott Rubin analyzes the motion of a snowball rolling without slipping down a snowy inclined which has an initial value of (5/7) g sin and a terminal value plane (making angle with the horizontal) and accreting ad- of (1/6) g sin . Equations (6) and (7) are plotted in Fig. 1 and ditional snow as it does so.1 have a similar shape to Rubin’s graphs of v (t) and a(t) because Equations (12) and (13) of Rubin’s paper are two coupled V = dr/dt rapidly approaches a constant value of (12 )–1/2 differential equations for v and r, but they cannot be solved (gk sin )1/2 so that r can then be linearly rescaled as t. analytically in terms of elementary functions of independent variable t. Instead, Rubin performs a numerical integration in 1. S. Rubin, “A variable-mass snowball rolling down a snowy a spreadsheet to find v (t), along with the translational acceler- slope,” Phys. Teach. 57, 150–151 (March 2019). ation a(t) of the snowball’s center of mass. Carl E. Mungan U.S. Naval Academy, Annapolis, MD

436 THE PHYSICS TEACHER ◆ Vol. 57, October 2019 letters

40 3 ) 2 30 (m/s velocity (m/s

a 2 v 20 ) 1 acceleration 10

0 0 0 0.2 0.4 0.6 0.8 1 distance rolled s (km)

Fig. 1. Graphs of the acceleration and velocity of the snowball as a function of the distance it has rolled downhill from Eqs. (1), (6), o and (7) using Rubin’s values of k = 5 cm, = 30 , and r0 = 2 m.

THE PHYSICS TEACHER ◆ Vol. 57, October 2019 437 Derivation of Four Equations in the Snowball Letter

Equation (2) The last equality in Eq. (12) of Rubin rearranges into 2πr υ = V (A1) k where V = dr / dt . Substituting Eq. (A1) into Eq. (13) of Rubin leads to 2 23k ⎛2πrV ⎞ 7 2π d(rV ) g sinθ − ⎜ ⎟ = (A2) 2 k 5 k dt 10πr ⎝ ⎠ which simplifies to 46π 14π 60π 14π g sinθ − V 2 = V 2 + rA ⇒ g sinθ − V 2 = rA (A3) 5k 5k ( ) 5k 5k where A = dV / dt . Multiplying through by 5k / π turns Eq. (A3) into Eq. (2) in the letter.

Equation (5) As one can verify by differentiation, the two integrals in Eq. (4) of the letter are

V 2 r 1/120 1/14 ln(C −60V ) ln(r) ⎛ C ⎞ ⎛ r ⎞ − = ⇒ ln⎜ ⎟ = ln⎜ ⎟ . (A4) 120 14 ⎝C 60V 2 ⎠ r r − ⎝ 0 ⎠ 0 0 Take the antilogarithm and then the 120th power of both sides to obtain

60/7 1 ⎛ r ⎞ = ⎜ ⎟ . (A5) 2 ⎜r ⎟ 1−60V / C ⎝ 0 ⎠ −1 Finally take the reciprocal of both sides, substitute in C = 5π gk sinθ , and then solve for V to get Eq. (5) in the letter.

Equation (6) Substitute Eq. (5) from the letter into Eq. (A1) above to obtain

2 ⎡ 60/7 ⎤ ⎛2πr r ⎞ gk sinθ ⎛r ⎞ υ = ⎜ 0 ⎟ ⎢1−⎜ 0 ⎟ ⎥ (A6) ⎜ ⎟ ⎢ ⎥ ⎝ k r0 ⎠ 12π ⎝ r ⎠ ⎣ ⎦ which simplifies to become Eq. (6) in the letter.

Equation (7) The acceleration of the snowball is

46/7 2r 46r0 ⎡ 60/7 ⎤ + dr dυ dυ gk sinθ ⎛r ⎞ π gr2 sinθ r2 7r53/7 a = =V = ⎢1−⎜ 0 ⎟ ⎥ 0 0 (A7) dt dr dr 12π ⎢ ⎝ r ⎠ ⎥ 3k 2 ⎡ 60/7 ⎤ ⎣ ⎦ ⎛ r ⎞ ⎛r ⎞ 2 ⎜ ⎟ ⎢1−⎜ 0 ⎟ ⎥ ⎝r0 ⎠ ⎢ ⎝ r ⎠ ⎥ ⎣ ⎦ using Eqs. (5) and (6) in the letter. This result simplifies to become Eq. (7) in the letter. A Variable-Mass Snowball Rolling Down a Snowy Slope Scott Rubin, The Berkeley Carroll School, Brooklyn, NY he stereotypical situation of a snowball picking up both snowball and is necessary for rotation to occur. Using the defi- mass and speed as it rolls without slipping down a hill nition of torque ( = rF), where r is the radius of the snowball, provides an opportunity to explore the general form of we have bothT translational and rotational versions of Newton’s second law through multivariable differential equations. With a few (4) reasonable assumptions, it can be shown that the snowball Since the radius, mass, and velocity are all changing, we need reaches a terminal acceleration. While the model may not be to use the general version of the rotational form of Newton’s completely physically accurate, the exercise and the resulting second law, equation are useful and accessible to students in a second year physics course, arguably. where L is angular momentum, I is rotational inertia, and is FN the angular velocity:

(5) Equation (4) becomes Ff

(6)

Fg = mg Next, substitute for a uniformly dense sphere:

(7) Fig. 1. Free-body diagram of a snowball as it rolls down an inclined plane of angle . The force of friction (Ff ) causes the Combining like terms and applying the relationship snowball to roll without slipping. yields

First, consider the free-body diagram of a spherical snow- (8) ball on an incline with angle (Fig. 1). Here, Fg is the down- ward gravitational force, FN is the normal force of the inclined But the rates of change of mass and radius are related. For a plane on the snowball, Ff is the force of friction, m is the mass uniformly dense sphere of density , of the snowball, and g is the acceleration of gravity. (9) The net external force acting on the snowball is then: which, upon differentiation yields Fnet = mg sin – Ff . (1) 2 . (10) For a thorough understanding of the situation, it is important to consider both the translational and the rotational aspects Substituting this relationship into Eq. (8) yields of the snowball’s motion. For the translational motion, use the . general form of Newton’s second law: (11) A reasonable assumption is the snowball’s radius increases by (2) a constant value k in every rotation:

And therefore, (12) (3) That gives a differential equation of motion: Note that Eq. (2) is only valid for a well-defined system upon which the net force is applied.1-3 In this case, the sys- (13) tem includes both the snowball and all the resting snowflakes waiting to be accreted. Since these waiting snowflakes remain This isn’t such an easy equation to solve since the radius is also at rest until they become part of the snowball, their velocities changing, but one more time derivative is instructive. Note and accelerations vanish and Eq. (2) should be valid. that the term g sin is constant. If we have Now consider the rotational aspect of the snowball’s mo- (14) tion. The force of friction, Ff, causes the torque acting on the

150 THE PHYSICS TEACHER ◆ Vol. 57, March 2019 DOI: 10.1119/1.5092471 70 4

60 3.5

) 3 50 ) 2.5 40 2 30

Speed (m/s 1.5 20 Acceleraon (m/s^2 1

10 0.5

0 0 0102030405060 0102030405060 Time (s) Time (s) Fig. 2. Speed-time graph for a snowball with k = 0.05 m, initial Fig. 3. Acceleration-time graph for a snowball with k = 0.05 m, speed = 0, initial radius = 2, and = 30o. initial speed = 0, initial radius = 2, and = 30o. 4-6 This suggests a terminal acceleration (aT) when number of droplets “swept out,” respectively. which, with substitution into Eq. (13), yields While a snowball, unlike a raindrop, doesn’t “sweep out” stationary snowflakes, and as it gets faster and larger its mass (15) clearly increases more quickly than a simple proportionality with time, it is instructive to follow the approach of Dick7 and This is a satisfying result for two reasons. First of all, it analyze Eq. (13) numerically. is relatable. Any student who has internalized the concept In the numerical model, I set initial values for k (in me- of terminal velocity should be able to extend their under- ters), , initial radius (in meters), and initial velocity (in m/s). standing to terminal acceleration. Secondly, it is intuitive. It I used time steps of 0.01 in an incremental version of Eqs. (12) wouldn’t make sense for a snowball to get slower as it rolled and (13): down a hill. Neither would it make sense for the snowball to accelerate as much as the usual value of a uniformly dense (18) sphere of constant radius rolling without slipping down an incline, namely: (19)

It also makes sense that the value k, the rate the radius in- Increasing k or decreasing r reduced the time it took to reach creases per rotation, doesn’t affect the final acceleration. It terminal acceleration, although, as indicated above, the termi- only affects the number of rotations before the terminal ac- nal acceleration remained the same for any size snowball on a celeration is reached. particular inclined plane. It is fair to note that the edges of the snowball may not gain The spreadsheet is available on request, but Figs. 2 and 3, mass at the same rate as the point of contact at its center, and respectively, show speed and acceleration for a representative hence the relationship snowball. Under these conditions, the snowball is very close to terminal acceleration by about 20 seconds. Acknowledgment may be a poor approximation. However, any model that I would like to thank Jefferson, Matteo, Julia, Jake, Ian, Mokhtar, maintains a spherical shape would require a similar approxi- Mia, Bryan, and Allen for asking the kinds of questions that mation. Nevertheless, it is instructive to apply this analysis to inspired this paper, as well as Bob Vitalo and Jane Moore for a rolling “snow-log”—a uniformly dense cylinder—to which giving me so much curricular freedom all these years. the approximation more clearly applies. Equation (8), for example, can be generalized to accom- References modate any rotating rigid body with a rotational inertia of I = 1. John Mallinckrodt, “F does not equal d(mv)/dt,” Phys. Teach. 48, mr2, where is a dimensionless constant between 0 and 1: 360 (Sept. 2010). 2. Martin Tiersten, “Force, momentum change, and motion,” Am. (16) J. Phys. 37, 82–22 (Jan. 1969). 3. J. Matolyak and G. Matous, “Simple variable mass systems: For a uniformly dense cylinder with length L and density Newton’s second law,” Phys. Teach. 28, 328–329 (May 1990). , m = r2 L and Since for a uniformly dense 4. K. S. Krane, “The falling raindrop: Variations on a theme of cylinder is 1/2, Eq. (13) becomes Newton,” Am. J. Phys. 49, 113–117 (Feb. 1981). 5. Alan D. Sokal, “The falling raindrop, revisited,” Am. J. Phys. 78, (17) 643–644 (June 2010). which yields a terminal acceleration of 6. Carl E. Mungan, “More about the falling raindrop,” Am. J. Phys. These results are reminiscent of the classic problem 78, 1421–1424 (Dec. 2010). in which a drop of rain that gains mass as it falls from rest 7. B. G. Dick, “On the raindrop problem,” Am. J. Phys. 54, 852–854 through a mist accelerates at a constant rate of g/2, g/4, and (Sept. 1986). g/7 if increasing mass is proportional to time, speed, and THE PHYSICS TEACHER ◆ Vol. 57, March 2019 151