Question:
A wheel of mass M is rolling down a ramp that is inclined at an angle θ with the horizontal. The ramp exerts a static frictional force on the wheel such that the wheel is rolling without slipping.
(a) (i) In the diagram below, draw and label the forces that act on the wheel as it rolls down the ramp.
(ii) Specify the force that rotates the wheel with respect to its center of mass? Explain your reasoning.
(b) Assume the frictional force exerts on the wheel is less than the maximum possible frictional force: the frictional force is 40% of the force due to the weight in the direction of the ramp. Derive the linear acceleration of the wheel.
(c) In a second experiment on the same ramp, the wheel is released from rest at the same time and the same height with a block of ice having the same mass M. (Assume the block is sliding down the ramp without friction.)
(i) Which object reaches the bottom of the ramp with the greatest speed? Explain your answer in terms of forces.
(ii) Explain your answer in terms of energy.
A wheel of mass M is rolling down a ramp that is inclined at an angle θ with the horizontal. The ramp exerts a static frictional force on the wheel such that the wheel is rolling without slipping.
(a) (i) In the diagram below, draw and label the forces that act on the wheel as it rolls down the ramp.
(ii) Specify the force that rotates the wheel with respect to its center of mass? Explain your reasoning.
(b) Assume the frictional force exerts on the wheel is less than the maximum possible frictional force: the frictional force is 40% of the force due to the weight in the direction of the ramp. Derive the linear acceleration of the wheel.
(c) In a second experiment on the same ramp, the wheel is released from rest at the same time and the same height with a block of ice having the same mass M. (Assume the block is sliding down the ramp without friction.)
(i) Which object reaches the bottom of the ramp with the greatest speed? Explain your answer in terms of forces.
(ii) Explain your answer in terms of energy.
Scoring guidelines:
(a) (i) gravitational force [1].
friction and normal force [1].
(ii) Explain that the frictional force exerts a torque with respect to the wheel’s center of mass. [1].
(b) An expression for the sum of the force components parallel to the ramp:
E.g. F‖ = Mg sin θ – Ff [1].
Indicate the frictional force is 0.4Mgsin θ and derive the acceleration in terms of g and θ: a = 0.6g sin θ [1].
(c) Explanation in terms of forces:
E.g. The wheel experiences a frictional force and the block has a greater net force and thus, a greater acceleration. [1]
Explanations in terms of energy:
E.g. The ice block and wheel lose the same amount of potential energy and gain the same amount of kinetic energy. For the ice block, the kinetic energy is purely translational; for the wheel, the kinetic energy is partly rotational. Thus, the block is faster because the wheel has lower translational kinetic energy. [1]
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q1.pdf)
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q1.pdf)
Possible answers:
(a) (i) There are three forces acting on the wheel: the gravitational force (mg), static frictional force (Ff), and normal force (N). The lengths of the arrows can be drawn carefully such that they indicate the relative magnitudes of the forces as shown below. For example, the magnitude of the normal force (N) should be lesser than the gravitational force (mg) because N = mg cos θ and thus, the length of the arrow representing N is shorter than mg.
(a) (ii) Static frictional force. With respect to the wheel’s center of mass, there is a static frictional force acting at the point of contact between the wheel and the ramp. This results in a non-zero torque about the wheel’s center of mass and the change in angular velocity. Alternatively, we can explain that the gravitational force on the wheel leads to a normal force and frictional force, and thus, a torque about the wheel’s center of mass. However, the velocity of various points on the wheel may vary with respect to the wheel’s center of mass and the point of contact. More importantly, the velocity of these points can be explained by the torque due to the static frictional force (about the center of mass) and gravitational force (about the point of contact) respectively as shown below.
(b) There are two forces acting on the wheel parallel to the ramp: static frictional force and the component of the wheel’s weight in the direction parallel to the ramp.
(The component of the wheel’s weight in the direction perpendicular to the ramp cancels with the normal force: Mg cos θ - Mg cos θ = 0)
By resolving the forces parallel to the ramp, ΣF = Mg sin θ – Ff
(The question states that the magnitude of the force of static friction exerted on the wheel is 40 percent of the magnitude of the force directed opposite to the frictional force. It means that the frictional force, Ff, is 0.4Mg sin θ.)
Therefore, Mg sin θ – 0.4Mg sin θ = 0.6Mg sin θ
By using Newton’s second law, a = ΣF/m = 0.6Mg sin θ/M = 0.6g sin θ
(c) (i) The ice-block reaches the bottom of the ramp with a greater speed than the wheel. This is because there is an opposing force on the wheel due to the static frictional force. The net force acting on the ice-block (parallel to the ramp) is greater than the wheel, and thus, the ice-block has a greater acceleration and greater speed.
(c) (ii) In both cases, there are transformations of gravitational potential energy into the same amount of final kinetic energy. However, the kinetic energy of the wheel is partly rotational and partly translational, whereas the kinetic energy of the ice-block is completely translational. Thus, the ice-block has more translational kinetic energy and greater speed.
Feynman’s insights or goofs?:
(a) In a sense, when a wheel is rolling down the ramp, it is the static frictional force that causes a change in the angular velocity of the wheel with respect to its center of mass. Importantly, there are two more forces acting on the wheel: the gravitational force on the wheel and the normal (reaction) force that is perpendicular to the ramp. As the angle of the ramp is θ, the resultant force on the wheel parallel to the ramp is mg sin θ. Furthermore, we can calculate the torque on the wheel about the wheel’s center of mass or the point of contact between the wheel and the ramp. However, if there is no gravitational force acting on the wheel, there would be no frictional force. The frictional force is also directly proportional to the normal force and it is dependent on the wheel’s weight. Therefore, it is not incorrect to state that the gravitational force can contribute to the change in the wheel’s angular velocity or torque.
In Feynman’s words, “[t]he torque is also often called the moment of the force. The origin of this term is obscure, but it may be related to the fact that ‘moment’ is derived from the Latin movimentum, and that the capability of a force to move an object (using the force on a lever or crowbar) increases with the length of the lever arm (Feynman et al., 1963, Section 18–2 Rotation of a rigid body).” Feynman has also emphasized that the value of the torque is dependent on the chosen axis of rotation. Interestingly, Leonardo da Vinci (1452–1519) envisaged the effective lever arm of a force and called it “the spiritual distance of the force” (French, 1971). In a similar sense, we can determine the torque about the point of contact between the wheel and the ramp as MgR sin θ in which R is the radius of the wheel and Rsin θ is the so-called spiritual distance. If Feynman knew about this term, he would have make fun of it. More importantly, the torque in rotating the wheel can be related to the wheel’s weight, Mg.
(b) Alternatively, we can derive an expression for the linear acceleration of the wheel’s center of mass by using the principle of conservation of energy as follows:
Change in potential energy = Translational work + Rotational work
(Let d be the distance travelled along the ramp and d sin θ is the change in height of the wheel.)
weight × change in height = (net force + frictional force) × distance travelled
Mg × d sin θ = Ma × d + 0.4Mg sin θ × d
By rearranging the terms, Ma × d = 0.6 Mgd sin θ
Therefore, a = 0.6g sin θ
(Alternatively, the rotational work can be calculated by using torque times angular displacement, τ × Δθ. Note that the torque is equal to the frictional force times the radius of the wheel, Ff × R, and the distanced traveled by the wheel is equal to the radius of the wheel times the angular displacement, d = RΔθ. Thus, the rotational work = τ × Δθ = [Ff × R] × Δθ = Ff × d)
According to Feynman, “[w]hen the object has rotated through a small angle Δθ, the work done, of course, is the component of force in the direction of the displacement times the displacement. In other words, it is only the tangential component of the force that counts, and this must be multiplied by the distance rΔθ. Therefore we see that the torque is also equal to the tangential component of force (perpendicular to the radius) times the radius (Feynman et al., 1963, section 18–2 Rotation of a rigid body).” That is, the rotational work that is done by the frictional force is not only equal to the torque multiplied by the angular displacement, τ × Δθ. Essentially, this rotational work can also be visualized as the frictional force multiplied by the distance, rΔθ.
(c) In the second experiment on the same ramp, an ice-block having the same mass M is released from rest at the same instant as the wheel. From a perspective of force, there is a component of the gravitational force, Mg sin θ, acting on both the ice-block and the wheel, however, there is an additional (opposing) force, μMg cos θ, acting on the wheel. Thus, the net force on the wheel is lesser because the static frictional force acts in the opposite direction. From a perspective of energy, the wheel has rotational kinetic energy in addition to its translational kinetic energy. As the wheel’s final translational kinetic energy is relatively lesser because of its rotation, it means that the wheel moves slower as compared to the ice-block that has purely translational kinetic energy. If the question states that the ice-block experiences the same amount of frictional (in this case, kinetic) force, it will reach the bottom of the ramp simultaneously as the wheel.
Leonardo da Vinci might have difficulties in answering this question correctly. Based on an analysis of his famous notebooks, da Vinci’s contributions to the development of the laws of friction include: (1) frictional force is independent of the surface contact area, (2) frictional force is directly proportional to the normal (contact) force, and (3) coefficient of frictional force is 0.25 (Pitenis, Dowson, & Sawyer, 2014). This coefficient of friction is reproducible under conditions of roughly cut and brusquely squared samples of dry wood. In general, the static and kinetic coefficients of frictional force vary from 0.3 to 0.6 for common materials and drop to about 0.15 when a lubricant is used. More importantly, ice-blocks that are close to 0 oC have static coefficients of friction about 0.05 and kinetic coefficients of friction from 0.02 to 0.04 (Mills, 2008). Thus, the static frictional force acting on the wheel should not be the same as the kinetic frictional force on the ice-block.
As a contrasting comparison, the kinetic frictional force prevents the ice-block from achieving a higher speed, whereas the static frictional force accelerates the wheel by rotating it. In short, it is not true that the effect of the static frictional force is to decelerate an object. Interestingly, Feynman mentions that “[t]here is another kind of friction, called dry friction or sliding friction, which occurs when one solid body slides on another. In this case, a force is needed to maintain motion (Feynman et al., section 12–2 Friction). However, one should not quote these words of Feynman and conclude that a driving force is always needed to overcome the frictional force such that an object can remain in motion. Conversely, it is possible that a static frictional force maintains the object’s motion or increases the velocity of the object.
Importantly, there are far more detailed discussions of frictional force in The Feynman Lectures on Physics as compared to many other physics textbooks. For instance, he mentions that “the coefficient of friction is only roughly a constant, and varies from place to place along the plane. The same erratic behavior is observed whether the block is loaded or not. Such variations are caused by different degrees of smoothness or hardness of the plane, and perhaps dirt, oxides, or other foreign matter. The tables that list purported values of μ for ‘steel on steel,’ ‘copper on copper,’ and the like, are all false, because they ignore the factors mentioned above, which really determine μ (Feynman et al., section 12–2 Friction).” Feynman disagrees with these coefficients of friction because they are dependent on the amount of impurities present. Currently, we can explain that the measured coefficients are dependent on the normal force, relative velocity, direction of motion relative to surface features, system stiffness, surface cleanliness, roughness, contact temperature, relatively humidity, lubricant properties, presence of loose particles, and experimental procedures (Blau, 2008).
References:
1. Blau, P. J. (2008). Friction science and technology: from concepts to applications (2nd ed.). Boca Raton: CRC press.
2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
3. French, A. (1971). Newtonian Mechanics. New York: W. W. Norton.
4. Mills, A. (2008). The coefficient of friction, particularly of ice. Physics Education, 43(4), 392.
5. Pitenis, A. A., Dowson, D., & Sawyer, W. G. (2014). Leonardo da Vinci’s friction experiments: An old story acknowledged and repeated. Tribology Letters, 56(3), 509-515.
