Classical-Mechanics · Unit 11 · Video 4 · Interactive Practice

The Tension That Spins: From Two Whirling Masses to a Whole Rope

IKey Formulas

Formula	Name	What you need
$\sum F_r = m a_r = -m r \omega^2$	Radial law of circular motion	$m$ , $r$ , $\omega$ , with $\hat{r}$ outward
$T_2 = 2 m_2 d\,\omega^2$	Outer string (two whirling masses)	$m_2$ , $d$ , $\omega$
$T_1 = d\,\omega^2\,(m_1 + 2 m_2)$	Inner string (two whirling masses)	$m_1$ , $m_2$ , $d$ , $\omega$
$\dfrac{dT}{dr} = -\dfrac{m}{L}\,\omega^2 r \;\Longrightarrow\; T(r) = \dfrac{m\omega^2}{2L}\left(L^2 - r^2\right)$	Spinning uniform rope	$m$ , $L$ , $\omega$ , and $T(L) = 0$

Key Insight: Tension accumulates inward — every segment must supply the inward pull for everything beyond it, so tension peaks at the shaft and dies at a free end.

IITwo Whirling Masses

Both masses circle at the same $\omega$ — why must the inner string pull harder than the outer one?

Inner radius r₁ (m) 1.0 Outer radius r₂ (m) 2.0

IIIFrom Beads to a Continuous Rope

The same total mass split into more beads turns the tension staircase into a smooth parabola.

Beads (N) 2

💡 The integration constant is fixed at the outer end: nothing lies beyond $r = L$ to pull back, so $T(L) = 0$ .

IVHalfway Out, But Not Half the Tension

At the midpoint of a spinning rope, how much of the shaft tension is still left?

Your guess: T at r = 2 m (N) 2.00

VQuiz Questions

Problem 1 · The Outer String

Given: two objects whirl about a vertical shaft at $\omega = 4$ rad/s — $m_1 = 3$ kg at radius $d = 0.5$ m and $m_2 = 1$ kg at radius $2d = 1.0$ m (massless strings, no gravity) — find the tension $T_2$ in the string between the two objects.

✅ Correct! Only

T_2

acts on the outer object, so

T_2 = m_2 (2d)\omega^2 = 1 \times 1.0 \times 16 = 16

❌ Check the radius. The outer object sits at

2d = 1.0

m, not at

d = 0.5

m — that factor of

2

doubles the answer.

❌ That is the inner string.

40

N is

T_1

, which must hold both masses; the outer string only has to hold

m_2

❌ Wrong mass. The outer object's free body contains only

m_2

—

m_1

never appears in

T_2

❌ Not quite. Apply the radial law to the outer object alone:

-T_2 = m_2\left(-(2d)\omega^2\right)

Show solution

Free body of the outer object: only one force, the tension $T_2$ pulling inward. With $\hat{r}$ outward, $a_r = -(2d)\omega^2$ :

-T_2 = m_2\,a_r = -m_2 (2d)\omega^2 \quad\Longrightarrow\quad T_2 = 2 m_2 d\, \omega^2

T_2 = 2(1)(0.5)(4)^2 = 2 \times 0.5 \times 16 = 16\ \text{N}

The tension is 16 N. Note that $m_1$ never enters — it lies inside this string, not beyond it.

Problem 2 · The Inner String (Common Pitfall)

Given: the same system — $m_1 = 3$ kg at $d = 0.5$ m, $m_2 = 1$ kg at $2d = 1.0$ m, $\omega = 4$ rad/s — find the tension $T_1$ in the string between the shaft and the inner object.

✅ Correct!

T_1 = d\omega^2(m_1 + 2m_2) = 0.5 \times 16 \times (3 + 2) = 40

N — the inner string supplies

m_1

's pull and transmits

T_2

❌ You accounted for $m_1$ only.

24\ \text{N} = m_1 d\omega^2

; the inner string must also deliver

T_2 = 16

N to the outer object, giving

24 + 16 = 40

❌ That is the outer string. The two strings are not under the same tension — the inner one carries strictly more.

❌ Check the radii. Only

m_2

sits at

2d

; the inner object is at

d

, so its term is

m_1 d\omega^2

, not

m_1 (2d)\omega^2

❌ Not quite. Newton's radial law on the inner object reads

T_2 - T_1 = -m_1 d\omega^2

Show solution

The inner object feels $T_2$ outward (toward $m_2$ ) and $T_1$ inward (toward the shaft), with $a_r = -d\omega^2$ :

T_2 - T_1 = -m_1 d\,\omega^2

Substituting $T_2 = 2m_2 d\omega^2$ from Problem 1:

T_1 = T_2 + m_1 d\,\omega^2 = 2 m_2 d\,\omega^2 + m_1 d\,\omega^2 = d\,\omega^2 (m_1 + 2 m_2)

T_1 = 0.5 (4)^2 (3 + 2 \cdot 1) = 8 \times 5 = 40\ \text{N}

Numerically: $T_1 = 16 + 24 = 40$ N, comfortably larger than $T_2 = 16$ N.

Problem 3 · Tension Along a Spinning Rope

Given: a uniform rope of mass $m = 3$ kg and length $L = 2$ m spins about a shaft at $\omega = 5$ rad/s with its far end free — find the tension at the shaft and at the rope's midpoint.

What is the tension at the shaft, $T(0)$ ?

What is the tension at $r = 1$ m?

✅ Correct!

T(r) = 18.75\,(4 - r^2)

N, so the shaft carries

75

N and the midpoint

56.25

N — three quarters, not one half.

❌ Check the shaft value. Set

r = 0

T(0) = \dfrac{m\omega^2 L}{2} = \dfrac{3 \times 25 \times 2}{2} = 75

N. Dropping the

\tfrac{1}{2}

gives

150

N; the zero belongs at the free end.

❌ The falloff is not linear.

T

depends on

r^2

T(1) = 18.75\,(4 - 1) = 56.25

N, which is

\tfrac{3}{4}

T(0)

— halfway out is not half the tension.

Show solution

With $T(L) = 0$ at the free end, integrating $\dfrac{dT}{dr} = -\dfrac{m}{L}\omega^2 r$ inward gives

T(r) = \frac{m\omega^2}{2L}\left(L^2 - r^2\right) = \frac{3(25)}{2(2)}\left(4 - r^2\right) = 18.75\,(4 - r^2)

At the shaft $(r = 0)$ :

T(0) = 18.75 \times 4 = 75\ \text{N} \qquad \left(= \frac{m\omega^2 L}{2}\right)

At the midpoint $(r = 1\ \text{m})$ :

T(1) = 18.75 \times 3 = 56.25\ \text{N}

The ratio is $\dfrac{L^2 - (L/2)^2}{L^2} = \dfrac{3}{4}$ — the profile is a parabola, not a straight line.

Problem 4 · Where Does the Tension Halve? (Transfer)

Given: a uniform rope of mass $m$ and length $L$ spinning at angular velocity $\omega$ with a free outer end — find the distance $r$ from the shaft at which the tension has fallen to exactly half its maximum value.

✅ Correct!

\dfrac{T(r)}{T(0)} = \dfrac{L^2 - r^2}{L^2} = \tfrac{1}{2}

gives

r = L/\sqrt{2} \approx 0.707L

— and it depends on neither

m

nor

\omega

❌ That assumes a straight-line falloff. Because

T \propto L^2 - r^2

, at

r = L/2

the tension is still

\tfrac{3}{4}

of the maximum, not

\tfrac{1}{2}

❌ Too far out. At

r = \tfrac{\sqrt{3}}{2}L

the bracket is

L^2 - r^2 = \tfrac{L^2}{4}

, so the tension is a quarter of the maximum.

❌ Not quite. Set

\dfrac{L^2 - r^2}{L^2} = \tfrac{1}{2}

and solve for

r

— remember to take the square root at the end.

Show solution

The tension is largest at the shaft, $T(0) = \dfrac{m\omega^2 L}{2}$ . Dividing the general profile by it:

\frac{T(r)}{T(0)} = \frac{\frac{m\omega^2}{2L}\left(L^2 - r^2\right)}{\frac{m\omega^2 L}{2}} = \frac{L^2 - r^2}{L^2}

Setting this equal to $\tfrac{1}{2}$ :

L^2 - r^2 = \frac{L^2}{2} \quad\Longrightarrow\quad r^2 = \frac{L^2}{2} \quad\Longrightarrow\quad r = \frac{L}{\sqrt{2}} \approx 0.707\,L

Every factor of $m$ and $\omega$ cancels, so the half-tension point sits at $0.707L$ for every uniform spinning rope — more than two thirds of the way out to the free end.

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