Constant speed still turns.

Velocity changes direction around a circle even if its magnitude is constant. The acceleration points toward the centre and has magnitude $v^{2}$ /r. Tangential velocity is perpendicular to this inward acceleration.

Centripetal force names the resultant inward force; it is not an extra force added to tension, gravity or friction. Identify which real forces provide that resultant. If the inward force disappears, the object initially travels along the tangent.

INTERACTIVE MODEL

Turn at constant speed

Radius / m4

READ THE GRAPH

Explore the relationship

Acceleration is inward; velocity is tangential. The net inward force supplies ma.

Centripetal acceleration / m s⁻²

Tap or focus a plotted point to read its values.

View plotted data as a table

Explore the relationship
Series	Speed / m s⁻¹	Centripetal acceleration / m s⁻²
Centripetal acceleration / m s⁻²	0	0
Centripetal acceleration / m s⁻²	0.166667	0.00694444
Centripetal acceleration / m s⁻²	0.333333	0.0277778
Centripetal acceleration / m s⁻²	0.5	0.0625
Centripetal acceleration / m s⁻²	0.666667	0.111111
Centripetal acceleration / m s⁻²	0.833333	0.173611
Centripetal acceleration / m s⁻²	1	0.25
Centripetal acceleration / m s⁻²	1.16667	0.340278
Centripetal acceleration / m s⁻²	1.33333	0.444444
Centripetal acceleration / m s⁻²	1.5	0.5625
Centripetal acceleration / m s⁻²	1.66667	0.694444
Centripetal acceleration / m s⁻²	1.83333	0.840278
Centripetal acceleration / m s⁻²	2	1
Centripetal acceleration / m s⁻²	2.16667	1.17361
Centripetal acceleration / m s⁻²	2.33333	1.36111
Centripetal acceleration / m s⁻²	2.5	1.5625
Centripetal acceleration / m s⁻²	2.66667	1.77778
Centripetal acceleration / m s⁻²	2.83333	2.00694
Centripetal acceleration / m s⁻²	3	2.25
Centripetal acceleration / m s⁻²	3.16667	2.50694
Centripetal acceleration / m s⁻²	3.33333	2.77778
Centripetal acceleration / m s⁻²	3.5	3.0625
Centripetal acceleration / m s⁻²	3.66667	3.36111
Centripetal acceleration / m s⁻²	3.83333	3.67361
Centripetal acceleration / m s⁻²	4	4
Centripetal acceleration / m s⁻²	4.16667	4.34028
Centripetal acceleration / m s⁻²	4.33333	4.69444
Centripetal acceleration / m s⁻²	4.5	5.0625
Centripetal acceleration / m s⁻²	4.66667	5.44444
Centripetal acceleration / m s⁻²	4.83333	5.84028
Centripetal acceleration / m s⁻²	5	6.25
Centripetal acceleration / m s⁻²	5.16667	6.67361
Centripetal acceleration / m s⁻²	5.33333	7.11111
Centripetal acceleration / m s⁻²	5.5	7.5625
Centripetal acceleration / m s⁻²	5.66667	8.02778
Centripetal acceleration / m s⁻²	5.83333	8.50694
Centripetal acceleration / m s⁻²	6	9
Centripetal acceleration / m s⁻²	6.16667	9.50694
Centripetal acceleration / m s⁻²	6.33333	10.0278
Centripetal acceleration / m s⁻²	6.5	10.5625
Centripetal acceleration / m s⁻²	6.66667	11.1111
Centripetal acceleration / m s⁻²	6.83333	11.6736
Centripetal acceleration / m s⁻²	7	12.25
Centripetal acceleration / m s⁻²	7.16667	12.8403
Centripetal acceleration / m s⁻²	7.33333	13.4444
Centripetal acceleration / m s⁻²	7.5	14.0625
Centripetal acceleration / m s⁻²	7.66667	14.6944
Centripetal acceleration / m s⁻²	7.83333	15.3403
Centripetal acceleration / m s⁻²	8	16
Centripetal acceleration / m s⁻²	8.16667	16.6736
Centripetal acceleration / m s⁻²	8.33333	17.3611
Centripetal acceleration / m s⁻²	8.5	18.0625
Centripetal acceleration / m s⁻²	8.66667	18.7778
Centripetal acceleration / m s⁻²	8.83333	19.5069
Centripetal acceleration / m s⁻²	9	20.25
Centripetal acceleration / m s⁻²	9.16667	21.0069
Centripetal acceleration / m s⁻²	9.33333	21.7778
Centripetal acceleration / m s⁻²	9.5	22.5625
Centripetal acceleration / m s⁻²	9.66667	23.3611
Centripetal acceleration / m s⁻²	9.83333	24.1736
Centripetal acceleration / m s⁻²	10	25

Probe along the horizontal axis0Centripetal acceleration / m s⁻²: 0

Explore: Double speed at the same radius and compare centripetal acceleration.

v=\frac{2\pi r}{T}\qquad a_c=\frac{v^2}{r}\qquad F_{\rm net}=\frac{mv^2}{r}

WORKED EXAMPLE

A 0.50 kg object moves at 4 $\mathrm{m}\,\mathrm{s}^{-1}$ in a circle of radius 2 m.

ac = $4^{2}$ /2 = 8 $\mathrm{m}\,\mathrm{s}^{-2}$ .
Inward net force = 0.50 × 8 = 4 N.

Assumed knowledge

Circular geometry and Newton’s second law.

Learning checkpoints & source

YOUR LEARNING CHECKPOINT

Distinguish speed from velocity in circular motion.
Find centripetal acceleration and net force.

QCAA Physics · Unit 3 · Gravity and motion ↗

READY TO TRY IT?

Constant speed still turns.

Turn at constant speed

Explore the relationship

A 0.50 kg object moves at 4 m s−1\mathrm{m}\,\mathrm{s}^{-1}ms−1 in a circle of radius 2 m.

Put the idea to work.

A 0.50 kg object moves at 4 $\mathrm{m}\,\mathrm{s}^{-1}$ in a circle of radius 2 m.