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Motion in a Plane - Uniform Circular Motion

Grade 11CBSEPhysics

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

🔑Concepts

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Uniform Circular Motion (UCM) is defined as the motion of an object traveling along a circular path at a constant speed. Although the speed is constant, the velocity is not, because the direction of motion changes continuously.

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The angular displacement θ\theta is the angle subtended by the radius vector at the center of the circular path. It is measured in radians (radrad).

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Angular velocity ω\omega is the rate of change of angular displacement. It is a vector quantity with the unit rad/srad/s.

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The linear velocity vv of the particle is always tangential to the circular path at any given point.

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In UCM, there is a constant acceleration acting towards the center of the circle called Centripetal Acceleration (aca_c). It is responsible for changing the direction of the velocity vector.

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The time period TT is the time taken by the object to complete one revolution, while frequency ν\nu is the number of revolutions completed per unit time: ν=1T\nu = \frac{1}{T}.

📐Formulae

ω=dθdt=2πT=2πν\omega = \frac{d\theta}{dt} = \frac{2\pi}{T} = 2\pi\nu

v=rωv = r\omega

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

ac=4π2ν2r=4π2rT2a_c = 4\pi^2 \nu^2 r = \frac{4\pi^2 r}{T^2}

a⃗c=−ω2r⃗\vec{a}_c = -\omega^2 \vec{r}

💡Examples

Problem 1:

A stone tied to the end of a string 1.5 m1.5 \text{ m} long is whirled in a horizontal circle with a constant speed. If the stone makes 1010 revolutions in 20 s20 \text{ s}, calculate the magnitude of the centripetal acceleration.

Solution:

Given: Radius r=1.5 mr = 1.5 \text{ m}, Number of revolutions n=10n = 10, Time t=20 st = 20 \text{ s}.

  1. Calculate frequency: ν=nt=1020=0.5 Hz\nu = \frac{n}{t} = \frac{10}{20} = 0.5 \text{ Hz}.
  2. Calculate angular velocity: ω=2πν=2×π×0.5=π rad/s\omega = 2\pi\nu = 2 \times \pi \times 0.5 = \pi \text{ rad/s}.
  3. Calculate centripetal acceleration: ac=ω2r=(π)2×1.5≈14.8 m/s2a_c = \omega^2 r = (\pi)^2 \times 1.5 \approx 14.8 \text{ m/s}^2.

Explanation:

The frequency is first determined to find the angular velocity. Since the speed is constant, we use the formula for centripetal acceleration involving ω\omega and rr.

Problem 2:

An aircraft executes a horizontal loop of radius 1 km1 \text{ km} with a steady speed of 900 km/h900 \text{ km/h}. Compare its centripetal acceleration with the acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution:

Given: r=1000 mr = 1000 \text{ m}, v=900 km/h=900×518=250 m/sv = 900 \text{ km/h} = 900 \times \frac{5}{18} = 250 \text{ m/s}.

  1. Centripetal acceleration: ac=v2r=25021000=625001000=62.5 m/s2a_c = \frac{v^2}{r} = \frac{250^2}{1000} = \frac{62500}{1000} = 62.5 \text{ m/s}^2.
  2. Ratio: acg=62.59.8≈6.38\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38.

Explanation:

The acceleration of the aircraft is approximately 6.386.38 times the acceleration due to gravity, highlighting the high 'g-force' experienced during such maneuvers.