Uniform Circular Motion
ICSE · Class 11 · Physics
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A particle moves on a circular path of radius 2 m with a constant speed of 4 m/s. What is the magnitude of its centripetal acceleration?
In uniform circular motion, which of the following quantities remains CONSTANT?
A particle completes one revolution in 4 seconds on a circular path. What is its angular velocity in rad/s?
The relationship between linear velocity (v), angular velocity (ω), and radius (r) is:
Sample Questions
A particle moves in a circle of radius 0.5 m with an angular velocity of 6 rad/s. What is its linear speed?
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3 m/s
Step 1: Use the formula v = rω. Step 2: r = 0.5 m, ω = 6 rad/s. Step 3: v = 0.5 × 6 = 3 m/s. Option B (12 m/s) is wrong — it incorrectly uses v = 2rω. Option C (6 m/s) is wrong — it ignores the radius factor. Option D (0.5 m/s) is wrong — it uses v = r/ω instead of v = rω. Always include units in your calculation to avoid such errors.
What is the direction of centripetal acceleration of a particle moving in a circle?
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Towards the centre of the circle
Step 1: In uniform circular motion, the velocity of the particle is always tangential (along the tangent to the circle). Step 2: The change in velocity (Δv) as the particle moves from one point to another is directed towards the centre of the circle. Step 3: Since acceleration = change in velocity / time, the acceleration is also directed towards the centre. Step 4: This is why it is called 'centripetal' acceleration — the word 'centripetal' means 'centre-seeking'. Option A is the direction of velocity, not acceleration. Option B is the direction of centrifugal force (pseudo force). Option D i
A car of mass 1000 kg moves on a circular road of radius 50 m at a speed of 10 m/s. What is the centripetal force acting on the car?
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2000 N
Step 1: Formula for centripetal force is F = mv²/r. Step 2: Given: m = 1000 kg, v = 10 m/s, r = 50 m. Step 3: F = 1000 × (10)² / 50 = 1000 × 100 / 50 = 100000/50 = 2000 N. Option A (200 N) is wrong — it uses m × v / r instead of m × v²/r. Option C (500 N) is wrong — it uses m × g/r approximately. Option D (5000 N) is wrong — it divides by the wrong value. Always remember to square the velocity in the centripetal force formula.
When a car moves on a level circular road, the centripetal force is provided by:
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Frictional force between tyres and road
Step 1: For a car moving on a level (horizontal) circular road, we need a force directed towards the centre of the circular path. Step 2: Gravity acts vertically downward and is balanced by the normal reaction of the road — neither provides the centripetal force. Step 3: The engine force propels the car forward (tangentially), not towards the centre. Step 4: The only horizontal force available is the friction between the tyres and the road surface, which acts towards the centre of the turn. Step 5: This is why on wet or icy roads, cars tend to skid — friction is reduced and cannot provide the
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