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Motion - Uniform Circular Motion

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Uniform Circular Motion (UCM) occurs when an object moves in a circular path with a constant speed.

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Even though the speed is constant, the velocity is constantly changing because the direction of motion changes at every point. Therefore, UCM is an accelerated motion.

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The acceleration in uniform circular motion is directed towards the center of the circle, known as centripetal acceleration.

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The distance covered in one complete revolution is equal to the circumference of the circle, given by 2πr2\pi r.

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Examples include the motion of the moon around the earth, a satellite in a circular orbit around the earth, and a cyclist moving on a circular track at a constant speed.

📐Formulae

v=2πrtv = \frac{2\pi r}{t}

Circumference=2πr\text{Circumference} = 2\pi r

Angular Displacement (θ)=Arc lengthr\text{Angular Displacement } (\theta) = \frac{\text{Arc length}}{r}

💡Examples

Problem 1:

An athlete completes one round of a circular track of diameter 200 m200\text{ m} in 40 s40\text{ s}. What will be the distance covered and the displacement at the end of 2 minutes 20 seconds2\text{ minutes } 20\text{ seconds}?

Solution:

  1. Radius r=2002=100 mr = \frac{200}{2} = 100\text{ m}.
  2. Total time t=2 min 20 s=(2×60)+20=140 st = 2\text{ min } 20\text{ s} = (2 \times 60) + 20 = 140\text{ s}.
  3. Number of rounds n=Total timeTime for one round=14040=3.5 roundsn = \frac{\text{Total time}}{\text{Time for one round}} = \frac{140}{40} = 3.5\text{ rounds}.
  4. Distance =n×2πr=3.5×2×227×100=2200 m= n \times 2\pi r = 3.5 \times 2 \times \frac{22}{7} \times 100 = 2200\text{ m}.
  5. Displacement: After 3.53.5 rounds, the athlete is at the diametrically opposite point from the start. Displacement =Diameter=200 m= \text{Diameter} = 200\text{ m}.

Explanation:

Distance is the total path length calculated by multiplying the number of laps by the circumference. Displacement is the shortest straight-line distance between the starting and ending positions.

Problem 2:

A cyclist goes around a circular track once every 2 minutes2\text{ minutes}. If the radius of the circular track is 105 metres105\text{ metres}, calculate his speed (vv). (Take π=227\pi = \frac{22}{7})

Solution:

Given: r=105 mr = 105\text{ m}, t=2 minutes=120 st = 2\text{ minutes} = 120\text{ s}. Using the formula v=2πrtv = \frac{2\pi r}{t} v=2×227×105120v = \frac{2 \times \frac{22}{7} \times 105}{120} v=2×22×15120v = \frac{2 \times 22 \times 15}{120} v=660120=5.5 m/sv = \frac{660}{120} = 5.5\text{ m/s}

Explanation:

The speed is calculated by dividing the total distance of one circumference by the time taken to complete one revolution.