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Solution. Answer: D. Justification: The radial acceleration for a car in a uniform circular motion is: v is the velocity of the car and r is the radius of ...
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F AC U L T Y O F E D U C AT I O NF AC U L T Y O F E D U C AT I O N D e p a r t m e n t o f C u r r i c u l u m a n d P e d a g o g y
F A C U L T Y O F E D U C A T I O N
Retrieved from: http://www.wonderwhizkids.com/index.php/physics/mechanics/circular-motion
A Ferrari is traveling in a uniform circular motion around a racetrack. What happens to the radial acceleration of the car if the velocity is doubled and the radius of the circle is halved?
A. It remains the same.
B. It increases by a factor of 2.
C. It increases by a factor of 4.
D. It increases by a factor of 8.
E. It decreases by a factor of 2.
Answer: D
Justification: The radial acceleration for a car in a uniform circular motion is:
v is the velocity of the car and r is the radius of the circular track
v r
Sonic is rolling towards a spring in order to quickly change the direction of his speed and make it around the loop. The mass of the giant blue hedgehog is 30 kg and he is rolling towards the spring at 20 m/s. The spring is massless (and therefore perfect), can compress 0.5 m and is attached to an immovable block.
You can assume that there is no friction between the ground and rolling Sonic.
What is the smallest the spring constant could be in order for Sonic to roll around the 30 m loop?
A. Sonic already has enough kinetic energy to complete the loop therefore the value of the spring constant is irrelevant. B. The value of the minimum spring constant is 71 kN/m. C. The value of the minimum spring constant is 88 kN/m. D. The value of the minimum spring constant is 119 kN/m. E. Sonic cannot make the loop regardless of the spring constant.
Remember: Treat Sonic like an indestructible point mass sliding along his trajectory. Acceleration due to gravity = 9.82 m/s^2.
If we notice that the loop is a case of circular motion we can figure out the minimum velocity required to make the loop by using the formula for radial acceleration:
The radius is half the diameter of 30 m. The minimum acceleration possible (and thus the minimum velocity possible) is the situation when the normal force provided by the loop, and acting on sonic, is zero. In that case the acceleration is only Sonic's acceleration due to gravity and thus we can find our velocity from the following formula:
Now our task is to figure out how the spring constant impacts Sonicās ability to get to the top of the loop with a velocity of vmin = 12.14 m/s
Since Sonic has to change direction on impact with the spring all of Sonicās energy after impact has to come from the elastic potential energy stored in the spring. Due to energy conservation we then know that the energy stored in the spring (ES) must be equal to the sum of Sonic's potential (EP) and kinetic (EK) energy at the top of the loop. Since we know the height of the loop and the minimum velocity required in order to maintain the circular path we can calculate Sonic's energy at the top of the loop:
Total Energy = Ek + EP = Ā½mvmin^2 + mgh
And since we know the spring can only compress 0.5 m we can use the formula for elastic potential energy of a spring:
ES = Ā½kx^2 (where k is the spring constant and x is the amount of compression)
Now we can equate the two expressions together to solve for our spring constant:
Notice that we did not need Sonicās initial velocity to solve this problem. Sonic loses a vast majority of his initial kinetic energy by slamming into the fully compressed spring before being launched in the opposite direction.
Answer A is incorrect because it forgets that the direction of Sonic's motion needs to be changed before his kinetic energy can be used to complete the loop.
Answer B is incorrect because it fails to account for Sonic's kinetic energy at the top of the loop.
Answer D is incorrect because it uses Sonicās initial velocity to solve for the spring constant k.
Answer E is incorrect due to the indestructible nature of Sonic and his surroundings. Even though Sonic completely compresses the spring and slams to a halt, the stored energy in the spring is still available to him and could propel him around the loop.
Answer: B
Justification: This is a 2D kinematics problem involving circular motion. We can start solving the problem by looking at the two different positions of the rider, where position 1 is at the top of the ferris wheel and position 2 is at the bottom of the ferris wheel:
1
2
We know that in each location the force of gravity F = mg acts on the rider in the downwards direction. We also know that radial acceleration ar is always directed towards the center of the circle, and therefore the force due to radial acceleration ( mar ) for position 1 is directed downwards, while for position 2 it is directed upwards. In both cases, m stands for the mass of the rider.
We can use this information to look at the normal force acting on the rider in each position. To do this we can draw two free body diagrams:
Position 1
N 1
mg
ar
Position 2
N 2
mg
ar
+ve direction
Using Newton's second law: Position 1: N 1 ā mg = m (-ar) ļ N 1 = mg ā mar = m(g ā ar) Position 2: N 2 ā mg = m(+ar) ļ N 2 = mg + mar = m(g + ar)
Given your results from the previous question, calculate N 1 and N 2 , the normal force exerted on the rider at position 1 (top of the Ferris wheel) and position 2 (bottom of the Ferris wheel).
The diameter of the Westcoast Wheel is 26 m.
The rider weighs 100 kg, acceleration due to gravity is 9.8 m/s^2.
Hint: You will need to measure the time it takes for the Ferris wheel to complete 1 revolution (for now please use 30 s).
A. N 1 = 7.5 x 10^5 N and N 2 = 12.1 x 10^5 N B. N 1 = 750 N and N 2 = 1210 N C. N 1 = 9.23 x 10^5 N and N 2 = 10.37 x 10^5 N D. N 1 = 923 N and N 2 = 1037 N E. N 1 = 1037 N and N 2 = 923 N
Answer: D
Justification: This question can be answered without any calculations. If we evaluate the options we can see that only D makes sense:
The options given in A & C result from unit conversion errors (using grams instead of kg) and yield enormous numbers which would not be rational.
Option E reverses the information so that the rider feels a greater normal force at the top of the ride, which is contrary to the situation.
Options B & D are the only two "realistic" choices. We are dealing with 980N for the person +/- the force felt at the top or bottom and remember that Ferris wheels usually move quite slowly so it is difficult to feel these changes in forces.
Option B would result in about 25% change in force, which is very unrealistic for a Ferris wheel.
Option D is much more realistic given the conditions of this problem.
