A car is traveling over the top of a hill, which we will model to be a piece of a circular arc of radius R. For each case below, draw careful to-scale motion maps and use them to construct the velocity vectors at the following three points: the top of the hill; equal time intervals on either side of the hill. Use the latter two velocity vectors to construct the average acceleration vector a at the top of the hill. Evaluate and comment on the direction and relative magnitude of a for the following cases:
(a) The car is moving at constant speed v over a hill of radius R.
(b) The car is moving at constant speed 2v. Quantitatively, how does the magnitude a change compared to that in part (a)?
(c)The car is moving at constant speed v over a hill of radius R/2. Quantitatively, how does a now compare to that obtained in part (a)?
(d)Brownie points:From your results of parts (a)-(c), obtain an expression for the magnitude of the acceleration for the case of uniform circular motion (constant speed). Hint: what variables could the acceleration depend on? There is only one possible combination of those variables that will give you the right units!And your graphical constructions in parts (a)-(c) should confirm the functional dependences.
(e)The car speeds up as it goes over the top of the hill; the car slows down as it goes over the top of the hill. Discuss your results for this non-uniform circular motion in the context of the “superposition principle:” the idea that a seemingly complex motion (accelerating over a hill) can be broken down into the superposition (addition) of two simpler motions – perhaps better described here as a decomposition.
To answer these questions, we need to understand the motion of a car traveling over the top of a hill. We will use motion maps and vectors to analyze the velocity and acceleration at different points.
1. Case (a): The car is moving at constant speed v over a hill of radius R.
To construct the velocity vectors, draw a motion map with arrows representing the direction and magnitude of the velocity at different points. At the top of the hill, the velocity vector points horizontally because that is the direction of motion. On either side of the hill, the velocity vectors point downwards, indicating that the car is moving downwards.
To calculate the average acceleration vector at the top of the hill, take the difference between the velocity vectors on either side of the top of the hill and divide by the time interval. The average acceleration vector will point downwards due to the change in velocity.
The magnitude of the acceleration in this case is equal to the gravitational acceleration, g. The direction is downward.
2. Case (b): The car is moving at constant speed 2v.
The velocity vectors at the top and on either side of the hill will have the same directions as in case (a), but their magnitudes will be doubled because the car is moving at a higher speed.
Therefore, the average acceleration vector at the top of the hill will have the same direction as in case (a), but its magnitude will be doubled as well.
3. Case (c): The car is moving at constant speed v over a hill of radius R/2.
The velocity vectors at the top and on either side of the hill will have the same directions as in case (a), but their magnitudes will be halved because the radius of the hill is halved.
Therefore, the average acceleration vector at the top of the hill will have the same direction as in case (a), but its magnitude will be halved as well.
4. Case (d): Obtaining an expression for the magnitude of the acceleration in uniform circular motion.
From the results of parts (a)-(c), we can observe that the magnitude of the acceleration depends only on the radius of the hill (R). Therefore, the magnitude of the acceleration in uniform circular motion is directly proportional to the radius (R). Mathematically, it can be represented as a = v^2 / R, where v is the velocity of the car.
5. Case (e): Non-uniform circular motion with speeding up or slowing down.
In the context of the superposition principle, we can decompose the motion into two simpler motions: circular motion and accelerating or decelerating motion. The circular motion component will have the same acceleration as in case (d), i.e., a = v^2 / R. The additional accelerating or decelerating motion will have its own acceleration vector. The resulting acceleration vector will be the vector sum of these two components.
Depending on whether the car is speeding up or slowing down, the direction of the acceleration vector will change accordingly. The magnitude of the acceleration vector will be the sum of the magnitudes of the two components.
The graphical constructions in cases (a)-(c) can help confirm the functional dependences and illustrate the resultant acceleration vector in case (e).