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Grade 11 Science · Physics

Circular motion and universal gravitation

Use this for “A 0.5 kg ball on a 1.2 m string swings in a circle at 6 m/s. What centripetal force does the string supply?”

Diagram

r = 1.2 mv = 6 m/sac = 30 m/s²
Centripetal acceleration: v²/r = 6²/1.2 = 30 m/s²
Centripetal force: mv²/r = 0.5 × 30 = 15 N
T = 2πr/v = 2π × 1.2/6 = 1.26 s
The string’s pull toward the center supplies it; let go and the ball flies off along the tangent.
This is an example. Type your own values to start a new problem.
Fc (N)
=
m (kg)
×
(
v (m/s)
)
2
r (m)
ac
Centripetal acceleration
calculated · m/s²
T
Period
calculated

Formulas

ac
=
v
2
/
r
30
=
6
2
/1.2
Fc
=
m
×
ac
15 = 0.5 × 30
T
=
2π
×
r
v
1.26 = 2π × 1.2/6

Assumptions

•
The ball goes round a flat circle at a constant speed: only its direction changes.
•
The net force (here the string’s pull) points to the center: Fc = mv²/r.
•
Let go, it moves off in a straight line along the tangent, not outward.
•
For a moon or a planet, far bigger circles, use “A satellite in orbit”.

Step-by-step

WE KNOW
m = 0.5 kg, r = 1.2 m, v = 6 m/s
FIND
centripetal acceleration (ac), centripetal force (Fc), period (T)
Step 1 · Find centripetal acceleration (ac)
Use ac = v²/r
Turning
takes
an
acceleration
toward
the
center:
v
2
over
r
.
ac
=
6
2
/1.2
ac = 36/1.2
ac
=
30
m/
s
2
Step 2 · Find centripetal force (Fc)
Use Fc = m × ac
Newton’s second law toward the center: the mass times the centripetal acceleration.
Fc = 0.5 × 30
Fc = 15 N
Step 3 · Find period (T)
Use T = 2π × r/v
One
turn
is
the
circumference,
2πr,
at
speed
v
.
T
=
2π
×
1.2/6
T
=
1.26
s
Check
Put the numbers back in. Both sides should match.
30
=
6
2
/1.2
✓
15 = 0.5 × 30 ✓
1.26 ≈ 2π × 1.2/6 ✓
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