SME
Grade 11 Science · Problem type
At the bottom of a swing
Use this for “A 2 kg ball on a 1.5 m string swings through the bottom at 3 m/s. What is the tension there?”
Diagram
Weight = m × g = 2 × 9.8 = 19.6 N
Net force = Tension − Weight = 31.6 − 19.6 = 12 N
ac = Net force/m = 12/2 = 6 m/s²
Speed: 3 m/s
Length of the string (radius): 1.5 m
This is an example. Type your own values to start a new problem.
T (N)
−
W (N)
=
m (kg)
×
(
v (m/s)
)
2
r (m)
The picture’s net force T − W = m × ac, with ac = v²/r, is this equation.
ac
Centripetal acceleration
calculated · m/s²
Fₙₑₜ
Net force
calculated
Formulas
ac
=
v
2
/
r
6
=
3
2
/1.5
W
=
m
×
9.8
19.6 = 2 × 9.8
Fₙₑₜ
=
m
×
ac
12 = 2 × 6
T
=
W
+
Fₙₑₜ
31.6 = 19.6 + 12
Assumptions
•
At the bottom the center of the circle is straight up, so the net force points up.
•
The tension must hold up the weight and also supply mv²/r: T = m(g + v²/r).
•
Up is +; the string is the radius r.
Step-by-step
WE KNOW
m = 2 kg, v = 3 m/s, r = 1.5 m
FIND
centripetal acceleration (ac), weight (W), net force (Fₙₑₜ), tension (T)
Step 1 · Find centripetal acceleration (ac)
Use ac = v²/r
Turning
takes
an
acceleration
toward
the
center:
v
2
over
r
.
ac
=
3
2
/1.5
ac = 9/1.5
ac
=
6
m/
s
2
Step 2 · Find weight (W)
Use W = m × 9.8
Earth pulls each kilogram with 9.8 N.
W
=
2
×
9.8
W
=
19.6
N
Step 3 · Find net force (Fₙₑₜ)
Use Fₙₑₜ = m × ac
The net force up is the mass times the centripetal acceleration.
Fₙₑₜ = 2 × 6
Fₙₑₜ = 12 N
Step 4 · Find tension (T)
Use T = W + Fₙₑₜ
The string holds the weight and adds the pull to the center.
T
=
19.6
+
12
T
=
31.6
N
Check
Put the numbers back in. Both sides should match.
6
=
3
2
/1.5
✓
19.6 = 2 × 9.8 ✓
12 = 2 × 6 ✓
31.6 = 19.6 + 12 ✓
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