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Circular motion – problems and solutions

Circular motion – problems and solutions

1. A 10-kg object moves in a circle at constant speed 4 m/s. If the radius of the circle is 0.5 meters, then :

1) The frequency of circle is 4/π Hz

2) The centripetal acceleration is 32 m.s-2

3) The centripetal force is 320 N

4) The period is 4π s.

Which are the true statements?

Known :

Mass of object (m) = 10 kg

The linear velocity (v) = 4 m/s

The radius of circle (r) = 0.5 meters

Solution :

1) The frequency of circle

v = 2 π r f

4 = 2 π (0.5) f

4 = π f

f = 4/π Hertz

2) The centripetal acceleration

as = v2 / r = 42 / 0,5 = 16 / 0,5 = 32 m/s2

3) The centripetal force

F = m as = (10)(32) = 320 N

4) Period

T = 1 : f = 1 : 4/π = 1 x π/4 = π/4

2. An object moving in a circle of radius 6 meters. If the object rounds 16 circles in 2 minutes, what is the linear velocity of the object.

Known :

Radius (r) = 6 meters

The angular velocity (ω) = 16 revolutions / 2 minutes = 8 revolutions / minutes = 8 revolutions / 60 seconds = 0.13 revolutions/second.

Wanted: The linear velocity (v)?

Solution :

v = r ω = (6 meters)(0.13 revolutions/second) = 0.8 meters/second

In radian :

1 revolution = 2π radian = 2(3.14) = 6.28 radian

The angular velocity = 8 (6.28) radians / 60 seconds = 50.24 radians / 60 seconds = 0.84 radians/second

v = r ω = (6 meters)(0.84 radians/second) = 5.04 radians/second.

3. An object with radius of 20/π cm rotates 4 times in 1 second. What is the linear velocity of the edge of object.

Known :

Radius (r) = 20/π cm = 20 / 3.14 cm = 6.4 cm = 0.064 meters

The angular velocity (ω) = 4 revolutions / 1 second = 4 revolutions / second.

1 revolution = (2)(3.14) radians = 6.28 radians

The angular velocity (ω) = (4)(6.28) radians/second = 25.12 radians/second

Wanted : The linear velocity of the edge of object (v)

Solution :

v = r ω = (0.064 meters)(25.12 radians/second) = 1.6 meters/second

4. An object moving in a circle at constant speed, the linear velocity of the object depends on…

See also  Optical instruments – problems and solutions

Solution :

The equation of the linear velocity of the circular motion :

Circular motion – problems and solutions 1

v = the linear velocity

d = 2πr = circumference

T = period = the time required for one complete revolution.

5. An object moving in a circle of radius 50 meters. If the angular speed of the object is 120 rpm, what are the time interval and the linear velocity of the object?

Known :

Radius (r) = 50 cm = 0.5 meters

The angular velocity (ω) = 120 rpm = 120 revolutions / 1 minute = 120 revolutions / 60 minutes = 2 revolutions / 1 second

1 revolution = 2π radian

The angular velocity (ω) = 2 (2π radians) / 1 second = 4π radians/second

Wanted : The time interval (T) and the linear speed (v)

Solution :

Period (T) :

Period is the time required for one complete revolution.

An object rotates two revolutions in 1 second = 1 revolutions per 0.5 seconds. Period = 0.5 second.

The linear velocity (v) :

v = r ω = (0.5 meters)(4π radians/second) = 2π meters/second.

  1. What is the difference between tangential speed and angular speed in circular motion?

    Answer: Tangential speed is the linear speed of a point on a rotating object and it indicates how fast the point is moving along its circular path. Angular speed, on the other hand, refers to how fast the angle changes as the object rotates. Tangential speed is typically measured in meters per second (m/s) whereas angular speed is usually measured in radians per second (rad/s).

  2. What is centripetal acceleration and how does it relate to circular motion?

    Answer: Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always points towards the center of the circle and is responsible for keeping the object in its circular path. The formula for centripetal acceleration is where is the tangential speed and is the radius of the circle.

  3. Why does an object in uniform circular motion have acceleration even if its speed is constant?

    Answer: While the magnitude (or size) of the speed remains constant in uniform circular motion, the direction of the velocity changes. Since acceleration is defined as a change in velocity, and velocity is a vector quantity with both magnitude and direction, any change in direction constitutes an acceleration. In this case, it’s centripetal acceleration.

  4. How does the required force to maintain circular motion relate to the mass of the object and the radius of the circle?

    Answer: The required force to maintain circular motion is given by the centripetal force formula: . As seen from the equation, the required force is directly proportional to the mass of the object and inversely proportional to the radius of the circle.

  5. Why do you feel pushed outwards when a car takes a sharp turn (e.g., in a roundabout)?

    Answer: This is due to the “fictitious” centrifugal force, which is a perceived force that acts outward on a body moving in a circular path. It’s not a real force in the sense of a push or pull, but rather an effect of inertia. Your body wants to move in a straight line (Newton’s first law), but the car’s walls or seatbelt exert a force to keep you moving in a circle. This creates the sensation of being pushed outwards.

  6. What role does friction play in circular motion, especially when a car is turning on a road?

    Answer: Friction between the tires and the road provides the necessary centripetal force that allows a car to turn. Without sufficient friction, the car would slide or skid, failing to follow its intended circular path.

  7. How does the centripetal force change if the radius of the circular path is halved but the speed remains the same?

    Answer: If the radius is halved and speed remains the same, the centripetal force will double, as centripetal force is inversely proportional to the radius.

  8. Why can’t we have centrifugal force without centripetal force in a circular motion?

    Answer: Centrifugal force is a reactive or “fictitious” force observed in a rotating frame of reference. It seems to push objects outward from the center of rotation. However, for an object to move in a circular path, there must be a real force acting towards the center, which is the centripetal force. Without centripetal force, there wouldn’t be circular motion to begin with, and therefore, no perception of centrifugal force.

  9. How is the gravitational force between Earth and the Moon responsible for the Moon’s circular orbit?

    Answer: The gravitational force between Earth and the Moon acts as the centripetal force that keeps the Moon in its orbit around Earth. Without this gravitational attraction, the Moon would move in a straight line instead of its circular (or more accurately, elliptical) orbit.

  10. If a string tied to a ball is shortened by half while twirling the ball in a circle at the same speed, how would the tension in the string change?

Answer: If the string’s length (which corresponds to the radius of the circular path) is halved while keeping the speed constant, the centripetal force required (and thus the tension in the string) will double. This is because centripetal force (and consequently, the tension for this scenario) is inversely proportional to the radius.

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