Calculating Centripetal Acceleration | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Acceleration from speed and radius

The inward acceleration becomes larger when the object moves faster.
At the same speed, a smaller circle needs a larger change in direction each second.

a=\dfrac{v^2}{r}

Symbols

$a$ = centripetal acceleration (m s⁻²)
$v$ = speed along the circle (m s⁻¹)
$r$ = radius of the path (m)

Use this equation for uniform circular motion when v and r are known.
The acceleration points towards the centre. The equation above gives its magnitude only.
Speed is squared. If v doubles and r stays fixed, a becomes four times as large.

Worked example

A direct acceleration calculation

An object moves at 4.0 m s⁻¹ in a circle of radius 2.0 m. Find its centripetal acceleration.

Substitute the speed and radius.
$a=\dfrac{v^2}{r}$
$a=\dfrac{4.0^2}{2.0}=8.0\text{ m s}^{-2}$

Answer

a=8.0\text{ m s}^{-2}\text{ towards the centre}

The source of a = v²/r

The velocity diagram gives the first relation for a short time.

\Delta v\approx v\Delta\theta

Divide by the time taken and use angular speed.

a=\dfrac{\Delta v}{\Delta t}\approx v\dfrac{\Delta\theta}{\Delta t}=v\omega

Replace angular speed with speed divided by radius.

a=v\omega=v\dfrac{v}{r}=\dfrac{v^2}{r}

Acceleration from angular speed

Replace speed with radius multiplied by angular speed when the question gives angular speed.

a=r\omega^2

Symbols

$a$ = centripetal acceleration (m s⁻²)
$r$ = radius of the path (m)
$\omega$ = angular speed (rad s⁻¹)

A third form uses speed and angular speed. It has appeared directly in Paper 4 (9702/41/O/N/25 Q1(b)(ii)).

a=v\omega

Values given	Equation to start with
speed v and radius r	use the speed–radius form
angular speed and radius	use the angular-speed form
speed and angular speed	use the mixed form
period T and radius r	first find angular speed from 2π/T

Worked example

A point on a rotating wheel

A point is 0.85 m from the centre of a wheel. The wheel turns at 140 rad s⁻¹. Find the speed and acceleration of the point (9702/42/O/N/24 Q1(a)).

Find the speed.
$v=r\omega$
$v=0.85\times140$
$v=120\text{ m s}^{-1}$
Find the acceleration.
$a=r\omega^2$
$a=0.85\times140^2$
$a=1.7\times10^4\text{ m s}^{-2}$

Answer

v=120\text{ m s}^{-1},\quad a=1.7\times10^4\text{ m s}^{-2}

Your turnQuick check

An object has centripetal acceleration 18 m s⁻² at a speed of 6.0 m s⁻¹. Find the path radius. Then state what happens to the acceleration if the speed doubles and the radius stays fixed.

Show worked answer

Rearrange the speed–radius form to find the radius.

r=\dfrac{v^2}{a}=\dfrac{6.0^2}{18}=2.0\text{ m}

Doubling the speed makes the acceleration four times as large, so the new acceleration is 72 m s⁻².

Common mistake

Use the radius of the object's circular path. Do not use a diameter or a distance measured from the edge of another object.

Acceleration from speed and radius

The inward acceleration becomes larger when the object moves faster.
At the same speed, a smaller circle needs a larger change in direction each second.

a=\dfrac{v^2}{r}

Symbols

$a$ = centripetal acceleration (m s⁻²)
$v$ = speed along the circle (m s⁻¹)
$r$ = radius of the path (m)

Use this equation for uniform circular motion when v and r are known.
The acceleration points towards the centre. The equation above gives its magnitude only.
Speed is squared. If v doubles and r stays fixed, a becomes four times as large.

Worked example

A direct acceleration calculation

An object moves at 4.0 m s⁻¹ in a circle of radius 2.0 m. Find its centripetal acceleration.

Substitute the speed and radius.
$a=\dfrac{v^2}{r}$
$a=\dfrac{4.0^2}{2.0}=8.0\text{ m s}^{-2}$

Answer

a=8.0\text{ m s}^{-2}\text{ towards the centre}

The source of a = v²/r

The velocity diagram gives the first relation for a short time.

\Delta v\approx v\Delta\theta

Divide by the time taken and use angular speed.

a=\dfrac{\Delta v}{\Delta t}\approx v\dfrac{\Delta\theta}{\Delta t}=v\omega

Replace angular speed with speed divided by radius.

a=v\omega=v\dfrac{v}{r}=\dfrac{v^2}{r}

Acceleration from angular speed

Replace speed with radius multiplied by angular speed when the question gives angular speed.

a=r\omega^2

Symbols

$a$ = centripetal acceleration (m s⁻²)
$r$ = radius of the path (m)
$\omega$ = angular speed (rad s⁻¹)

A third form uses speed and angular speed. It has appeared directly in Paper 4 (9702/41/O/N/25 Q1(b)(ii)).

a=v\omega

Values given	Equation to start with
speed v and radius r	use the speed–radius form
angular speed and radius	use the angular-speed form
speed and angular speed	use the mixed form
period T and radius r	first find angular speed from 2π/T

Worked example

A point on a rotating wheel

A point is 0.85 m from the centre of a wheel. The wheel turns at 140 rad s⁻¹. Find the speed and acceleration of the point (9702/42/O/N/24 Q1(a)).

Find the speed.
$v=r\omega$
$v=0.85\times140$
$v=120\text{ m s}^{-1}$
Find the acceleration.
$a=r\omega^2$
$a=0.85\times140^2$
$a=1.7\times10^4\text{ m s}^{-2}$

Answer

v=120\text{ m s}^{-1},\quad a=1.7\times10^4\text{ m s}^{-2}

Your turnQuick check

An object has centripetal acceleration 18 m s⁻² at a speed of 6.0 m s⁻¹. Find the path radius. Then state what happens to the acceleration if the speed doubles and the radius stays fixed.

Show worked answer

Rearrange the speed–radius form to find the radius.

r=\dfrac{v^2}{a}=\dfrac{6.0^2}{18}=2.0\text{ m}

Doubling the speed makes the acceleration four times as large, so the new acceleration is 72 m s⁻².

Common mistake

Use the radius of the object's circular path. Do not use a diameter or a distance measured from the edge of another object.