# Relationship between Angular Acceleration & Linear Acceleration

### Speed:

From the diagram, as the particle P moves round the circle once it sweeps through an angle = 360º. Where 360º = 2π radians, in time T seconds.

The time rate of change of angle with time (t) is called the angular velocity (ω).

ω = \( \frac{angle \: turned \: through \: the \: body}{time \: taken} \)

ω = \( \frac{θ}{t}\)

∴ θ = ωt

This is similar to:

distance = velocity x time

s = v x t

∴ v = \( \frac{s}{t}\) for linear motion;

The angle *θ* is measured in radians

since, 2π rad = 360º

from, ω = \( \frac{θ}{t}\)

angular velocity is measured in radians per second (rad/s)

When *θ* changes with time, the length of the arc PZ = s, also changes with time.

By definition, *θ* in radians = \( \frac{s}{r}\)

s = r*θ*

Let r = A = radius of the circle.

The angular velocity (ω) is given by:

ω = \( \frac{θ}{t} = \frac{s}{r} \: \times \: \frac{1}{t} \)

= \( \frac{s}{t} \times \frac{1}{r} \)

\( \frac{s}{t} \) = v, the linear velocity of the particle.

Therefore;

ω = v. \( \frac{1}{r}\)

V = ωr = ωA.

Linear speed is the product of the angular speed and the radius or amplitude of motion.

Where,

v is the linear velocity of the object that is moving in a circular path, measured in m/s.

r is the radius of the circular path in which the object moved round, measured in metres.

ω is the angular velocity of the object, measured in radian per second (rad/s).

When t = T, the period of SHM or the time for one cycle, *θ* = 360º or 2π rad

∴ ω = \( \frac{2 \pi}{T} \)

or

⇒ T = \( \frac{2 \pi}{\omega} \)

but f = \( \frac{1}{T} \)

∴ \( \frac{1}{T} = \frac{\omega}{2 \pi} \)

⇒ \( \scriptsize \omega = 2 \pi f \)

Where we have:** ω:** angular speed or angular frequency

*T**period*

**:****frequency**

*f:*### Example 1:

A loaded tube of 2.5kg mass placed in a liquid of density 9.5g/cm^{-3} was made to oscillate vertically when slightly pushed down. If it performs 50 oscillations in 35 seconds, calculate **(i)** periodic frequency.**(ii)** angular speed.

**Solution:**

**i.** \( \scriptsize \rho = 9.5 g/cm^{-3} \)

m = 2.5kg

number of oscillations = 50

t = 35s.

Period, T = Time taken for one oscillation

T = \( \frac{time \: taken \: for \: no. \: of \: oscillations}{no. \: of \: oscillations} \)

T = \( \frac{35}{50} \)

T = 0.7s

f = \( \frac{1}{T} \)

= \( \frac {1}{0.7} \)

f = 1.43 Hz

**ii.** Angular speed, ω = \( \frac{2 \pi}{T}\) = 2πf

= \( \scriptsize 2 \: \times \: 3.14 \: \times 1.43 \)

= 8.98rads^{-1}.

### Example 2:

A stone is tied to one end of an inelastic string and whirled round in a circular path of radius 30cm. If the stone makes 9 complete revolutions in 3 secs, find its angular and linear velocities during the period.

**Solution:**

1 rev = 2π

9 rev = 9 x 2π

= 18π

t = 3s

ω = \( \frac{θ}{t} \\= \frac{18 \pi}{3} \)

= 6π rads^{-1}.

v = ωr

= 6π x 30

= 180π ms^{-1}.

At any point, Q whose distance from the central point O is x, the linear velocity is given by

v = \( \scriptsize \omega \sqrt{A^2 \: -\: x^2} \)

Maximum velocity or the velocity at the equilibrium position, v_{max} is attained when x = 0

∴ v_{max} = ωA

The minimum velocity occurs at the extreme position of motion (x = A)

### Acceleration:

We can obtain an expression relating angular velocity and linear acceleration, using differential calculus.

From the equation of displacement of a particle;

x = A cos *θ*

But θ = ωt

∴ x = A cos ωt

differentiating with respect to t, we have

\( \frac{dy}{dx} \scriptsize = -\omega A \: sin \omega t \)But \( \frac{dx}{dt} \) = rate of change of displacement with time

and, rate of change of displacement with time = linear velocity, v

∴ \( \frac{dx}{dt} \scriptsize = v =\: – \omega A \: sin \omega t \)

acceleration = rate of change of velocity with time = \( \frac{dv}{dt} \)

∴ a = \( \frac{dv}{dt} \scriptsize = \omega^2 A \: cos \omega t \)

but x = A cos ωt

∴ acceleration, a = \(\scriptsize – \omega^2 x \)

From the equation, the linear acceleration is equal to the product of the square of the angular speed and displacement, x, of the particle from the centre of motion.

The Acceleration is proportional to the displacement and in the opposite direction of displacement which fulfils the basic condition of SHM.

### Relationship between Angular Acceleration & Linear Acceleration:

Angular acceleration (∝) is the time rate of increase in angular velocity (ω). It is measured in rads^{-2}.

If the angular velocity of a body changes uniformly from ω_{0} to ω_{t} in t seconds, angluar acceleration (∝) is given by:

but v = ωr or ω = \( \frac{v}{r} \)

∴ \( \scriptsize \propto \: = \left(\large \frac{\frac{v_t}{r} \: – \: \frac{v_0}{r}}{t} \right ) \)

⇒ \( \scriptsize \propto \: = \left( \normalsize \frac{v_t \: – \: v_0}{rt} \right ) \)

∴ \( \scriptsize \propto \: = \normalsize \frac{1}{r}\left( \frac{v_t \: – \: v_0}{t} \right ) \)

⇒ \( \scriptsize \propto \: = \normalsize \frac{1}{r}\scriptsize .a \)

⇒ \( \scriptsize \propto \: = \normalsize \frac{a}{r} \)

a = \( \scriptsize \propto r \)

where a is the linear acceleration of the body.

Linear acceleration is the product of angular acceleration (∝) and the radius, or the displacement of the particle, from its central position.

**Summary:** Uniform circular motion occurs when a point P moves on a circular route with a constant angular velocity. Its x-axis projection is subject to simple harmonic motion. This is comparable to the linear vertical motion of an oscillating mass projected onto a spring.

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