Test your knowledge with questions about uniform circular motion and clear your doubts with comments in the resolutions.
question 1
(Unifor) A carousel rotates evenly, making one full rotation every 4.0 seconds. Each horse performs uniform circular movement with a frequency in rps (revolution per second) equal to:
a) 8.0
b) 4.0
c) 2.0
d) 0.5
e) 0.25
Correct alternative: e) 0.25.
The frequency (f) of the movement is given in time units according to the division of the number of laps by the time taken to execute them.
To answer this question, just replace the statement data in the formula below.
If a lap is taken every 4 seconds, the frequency of movement is 0.25 rps.
See too: Circular motion
question 2
A body in MCU can make 480 turns in a time of 120 seconds around a circumference of radius 0.5 m. Based on this information, determine:
a) frequency and period.
Correct answers: 4 rps and 0.25 s.
a) The frequency (f) of the movement is given in time units according to the division of the number of laps by the time taken to execute them.
The period (T) represents the time interval for the movement to repeat itself. Period and frequency are inversely proportional quantities. The relationship between them is established through the formula:
b) angular velocity and scalar velocity.
Correct Answers: 8 rad/s and 4
m/s.
The first step in answering this question is to calculate the angular velocity of the body.
Scalar and angular velocity are related from the following formula.
See too: Angular Velocity
question 3
(UFPE) The wheels of a bicycle have a radius equal to 0.5 m and rotate with an angular velocity equal to 5.0 rad/s. What is the distance covered, in meters, by this bicycle in a time interval of 10 seconds.
Correct answer: 25 m.
To solve this question, we must first find the scalar velocity by relating it to the angular velocity.
Knowing that scalar velocity is given by dividing the displacement interval by the time interval, we find the distance covered as follows:
See too: Average Scalar Velocity
question 4
(UMC) On a circular horizontal track, with a radius equal to 2 km, an automobile moves with a constant scalar speed, whose module is equal to 72 km/h. Determine the magnitude of the centripetal acceleration of the car, in m/s2.
Correct answer: 0.2 m/s2.
As the question asks for centripetal acceleration in m/s2, the first step in solving it is to convert the radius and speed units.
If the radius is 2 km and knowing that 1 km is 1000 meters, then 2 km corresponds to 2000 meters.
To convert speed from km/h to m/s just divide the value by 3.6.
The formula for calculating centripetal acceleration is:
Substituting the values of the statement in the formula, we find acceleration.
See too: centripetal acceleration
question 5
(UFPR) A point in uniform circular motion describes 15 revolutions per second on a circumference of 8.0 cm in radius. Its angular velocity, its period and its linear velocity are, respectively:
a) 20 rad/s; (1/15) s; 280 π cm/s
b) 30 rad/s; (1/10) s; 160 π cm/s
c) 30 π rad/s; (1/15) s; 240 π cm/s
d) 60 π rad/s; 15 s; 240 π cm/s
e) 40 π rad/s; 15 s; 200 π cm/s
Correct alternative: c) 30 π rad/s; (1/15) s; 240 π cm/s.
1st step: calculate the angular velocity applying the data in the formula.
2nd step: calculate the period by applying the data in the formula.
3rd step: calculate the linear velocity by applying the data in the formula.
question 6
(EMU) About uniform circular motion, check whichever is correct.
01. Period is the amount of time it takes a mobile to make a complete turn.
02. The rotation frequency is given by the number of turns a mobile makes per unit of time.
04. The distance that a mobile in uniform circular motion travels when making a complete turn is directly proportional to the radius of its trajectory.
08. When a rover makes a uniform circular movement, a centripetal force acts on it, which is responsible for the change in the rover's velocity direction.
16. The magnitude of centripetal acceleration is directly proportional to the radius of its trajectory.
Correct answers: 01, 02, 04 and 08.
01. CORRECT When we classify the circular movement as periodic, it means that a complete revolution is always given in the same time interval. Therefore, period is the time it takes the mobile to make a complete turn.
02. CORRECT Frequency relates the number of laps to the time taken to complete them.
The result represents the number of laps per unit of time.
04. CORRECT When making a complete turn in the circular movement, the distance covered by a mobile is the measure of the circumference.
Therefore, the distance is directly proportional to the radius of its trajectory.
08. CORRECT In circular motion, the body does not follow a trajectory, as a force acts on it, changing its direction. The centripetal force acts by directing you towards the center.
Centripetal force acts on the velocity (v) of the mobile.
16. WRONG. The two quantities are inversely proportional.
The magnitude of the centripetal acceleration is inversely proportional to the radius of its trajectory.
See too: Circumference
question 7
(UERJ) The average distance between the Sun and Earth is about 150 million kilometers. Thus, the average speed of translation of the Earth relative to the Sun is approximately:
a) 3 km/s
b) 30 km/s
c) 300 km/s
d) 3000 km/s
Correct alternative: b) 30 km/s.
As the answer must be given in km/s, the first step to facilitate the resolution of the question is to put the distance between Sun and Earth in scientific notation.
As the trajectory is performed around the Sun, the movement is circular and its measurement is given by the perimeter of the circumference.
The translation movement corresponds to the trajectory made by the Earth around the Sun in a period of approximately 365 days, that is, 1 year.
Knowing that a day is 86,400 seconds, we calculate how many seconds there are in a year by multiplying by the number of days.
Passing this number to scientific notation, we have:
The translation speed is calculated as follows:
See too: Kinematics Formulas
question 8
(UEMG) On a trip to Jupiter, it is desired to build a spaceship with a rotational section to simulate, by centrifugal effects, gravity. The section will have a radius of 90 meters. How many revolutions per minute (RPM) should this section have to simulate Earth's gravity? (consider g = 10 m/s²).
a) 10/π
b) 2/π
c) 20/π
d) 15/π
Correct alternative: a) 10/π.
Calculation of centripetal acceleration is given by the following formula:
The formula that relates linear velocity to angular velocity is:
Replacing this relationship in the centripetal acceleration formula, we have:
Angular velocity is given by:
By transforming the acceleration formula we arrive at the relationship:
Replacing the data in the formula, we find the frequency as follows:
This result is in rps, which means rotations per second. Through the rule of three we find the result in revolutions per minute, knowing that 1 minute has 60 seconds.
question 9
(FAAP) Two points A and B are located respectively 10 cm and 20 cm from the axis of rotation of the wheel of a uniformly moving automobile. It is possible to say that:
a) The period of movement of A is shorter than that of B.
b) The frequency of movement of A is greater than that of B.
c) The angular velocity of movement of B is greater than that of A.
d) The angular velocities of A and B are equal.
e) The linear velocities of A and B have the same intensity.
Correct alternative: d) The angular velocities of A and B are equal.
A and B, although at different distances, are located on the same axis of rotation.
As period, frequency and angular velocity involve the number of turns and the time to execute them, for points A and B these values are equal and, therefore, we discard alternatives a, b and c.
Thus, alternative d is correct, as observing the angular velocity formula , we came to the conclusion that as they are on the same frequency, the speed will be the same.
The alternative e is incorrect, as the linear velocity depends on the radius, according to the formula , and the points are situated at different distances, the speed will be different.
question 10
(UFBA) A spoke wheel R1, has linear velocity V1 at points located on the surface and linear velocity V2 at points 5 cm from the surface. being V1 2.5 times greater than V2, what is the value of R1?
a) 6.3 cm
b) 7.5 cm
c) 8.3 cm
d) 12.5 cm
e) 13.3 cm
Correct alternative: c) 8.3 cm.
On the surface, we have linear velocity
At points 5 cm farther from the surface,
The points are located on the same axis, hence the angular velocity () it's the same. How V1 is 2.5 times larger than v2, the speeds are related as follows: