Classical-Mechanics · Unit 18 · Video 2 · Interactive Practice

Why Energy Alone Can’t Clear the Loop

IKey Formulas

Formula	Name	What it gives you
$\tfrac{1}{2}kx^2 = 2mgR + \tfrac{1}{2}mv_{top}^2$	Energy conservation, spring $\to$ top	One equation, two unknowns ( $x$ and $v_{top}$ )
$-mg - N = -\dfrac{mv_{top}^2}{R}$	Newton's Second Law, radial, at the top	Both forces point to the center
$v_{top} = \sqrt{3gR}$	Speed at the top when $N = 2mg$	Mass cancels: only $g$ and $R$ survive
$x = \sqrt{\dfrac{7mgR}{k}}$	Spring compression	The two principles combined

Key Insight: Energy conservation prices the two states but leaves $x$ and $v_{top}$ tangled in a single equation. Newton's Second Law at the one point where the track's push is specified supplies the second, independent equation.

IIVisualization 1 — Where the Spring's Energy Goes

The spring's stored energy buys exactly two things: the climb to height $2R$ and the kinetic energy left at the top.

Journey spring

IIIVisualization 2 — The Track's Push Sets the Speed

At the top both forces point to the center, so a harder push from the track means a faster block.

N / mg 2.00

💡 Push the track's force below $N = 0$ and there is nothing left to hold the block on the circle: it leaves the track and flies as a projectile, so $\sqrt{gR}$ is the slowest speed a loop of radius $R$ can carry.

IVVisualization 3 — Two Principles, One Compression

Both principles describe the same $\tfrac{1}{2}mv_{top}^2$ , and equating them fixes the compression $x$ .

Step 1

Step 1 — Energy conservation, part (a)

Frictionless track, no air resistance, so

W_{nc} = 0

and

E_f = E_i

2mgR + \tfrac{1}{2}mv_{top}^2 = \tfrac{1}{2}kx^2 \quad\Longrightarrow\quad \tfrac{1}{2}mv_{top}^2 = \tfrac{1}{2}kx^2 - 2mgR

Two unknowns,

x

and

v_{top}

, in one equation — energy alone stops here.

Step 2 — Newton's Second Law at the top, part (b)

Taking

+y

up, both

mg

and

N

point down, toward the center:

-mg - N = -\frac{mv_{top}^2}{R}, \qquad N = 2mg \;\Longrightarrow\; 3mg = \frac{mv_{top}^2}{R}

Multiplying by

\tfrac{R}{2}

turns the force law into an energy statement:

\tfrac{1}{2}mv_{top}^2 = \tfrac{3}{2}mgR \qquad (v_{top} = \sqrt{3gR})

Step 3 — One quantity, two expressions

\tfrac{1}{2}kx^2 - 2mgR = \tfrac{3}{2}mgR

Step 4 — Read the budget

\tfrac{1}{2}kx^2 = \tfrac{3}{2}mgR + 2mgR = \tfrac{7}{2}mgR

Of the

\tfrac{7}{2}mgR

the spring stores,

2mgR

pays for the climb to height

2R

and

\tfrac{3}{2}mgR

is still kinetic at the top.

Step 5 — Solve for the compression

x^2 = \frac{7mgR}{k} \quad\Longrightarrow\quad x = \sqrt{\frac{7mgR}{k}}

Stiffer

k

gives a smaller

x

; larger

m

R

gives a larger

x

VQuiz Questions

Problem 1 · Speed at the Top

Given: a block of mass $m$ passes the top of a vertical loop of radius $R$ , where the track pushes on it with a normal force of magnitude $N = 2mg$ — find the speed $v_{top}$ .

✅ Correct! The weight and the normal force add to

3mg

of inward force, so

v_{top}^2 = 3gR

— set by

g

and

R

alone.

❌ Gravity is missing. At the top the weight also points toward the center, so the net radial force is

mg + N = 3mg

, not

N = 2mg

❌ That is the upward- $N$ answer. Drawing

N

upward gives

N - mg = mg

and

v^2 = gR

. A track can only push, and from the top of the loop pushing means pushing down.

❌ One algebra step short. From

\tfrac{1}{2}mv_{top}^2 = \tfrac{3}{2}mgR

, multiply by

\tfrac{2}{m}

v_{top}^2 = 3gR

, not

\tfrac{3}{2}gR

❌ Not quite. Write the radial equation with both forces pointing at the center, then substitute

N = 2mg

Show solution

At the top of the loop the center of the circle is below the block, so the centripetal acceleration points down, and so do both forces. With $+y$ up:

-mg - N = -\frac{mv_{top}^2}{R}

Substituting the given condition $N = 2mg$ :

-mg - 2mg = -\frac{mv_{top}^2}{R} \quad\Longrightarrow\quad 3mg = \frac{mv_{top}^2}{R}

The mass cancels and $v_{top}^2 = 3gR$ , so

v_{top} = \sqrt{3gR}

Equivalently $\tfrac{1}{2}mv_{top}^2 = \tfrac{3}{2}mgR$ — the kinetic energy at the top, fixed by forces alone, with no reference to the spring.

Problem 2 · The Slowest Legal Pass

Given: the same loop of radius $R$ , but with no condition imposed on $N$ — find the smallest speed at the top for which the block is still in contact with the track.

✅ Correct! Contact is lost the moment

N

would have to go negative, so the limiting case is

N = 0

and gravity alone supplies

mv^2/R

❌ That is energy thinking only. Merely arriving at the top is not enough: with

v = 0

the required centripetal force is zero while gravity still pulls with

mg

, so the block has already left the track.

❌ Check the radius.

\sqrt{2gR}

comes from writing

a = v^2/(2R)

with the diameter. The centripetal acceleration uses the radius:

a = v^2/R

❌ That is the $N = 2mg$ case. Here nothing is imposed on

N

, so the limit is the smallest push a track can give:

N = 0

❌ Not quite. Ask what the smallest normal force a track can exert is, then put that value into the radial equation.

Show solution

A track can push but never pull, so $N \ge 0$ . The radial equation at the top with $+y$ up is

-mg - N = -\frac{mv^2}{R} \quad\Longrightarrow\quad v^2 = \frac{(mg + N)R}{m}

The speed is smallest when $N$ is smallest, i.e. $N = 0$ :

v_{min}^2 = gR \quad\Longrightarrow\quad v_{min} = \sqrt{gR}

At exactly this speed gravity alone bends the block around the circle. Any slower and the required centripetal force is less than $mg$ , which the track cannot arrange — the block falls away from the loop before reaching the top.

Problem 3 · A Stiffer Condition

Given: the same spring ( $k$ ), block ( $m$ ) and frictionless loop ( $R$ ) as in the video, but the catch is set so that the normal force at the top is $N = 4mg$ — find the kinetic energy at the top and the compression $x$ .

Kinetic energy at the top?

Compression of the spring?

✅ Correct! A firmer push at the top costs more stored energy:

\tfrac{1}{2}kx^2 = 2mgR + \tfrac{5}{2}mgR = \tfrac{9}{2}mgR

, so

x = 3\sqrt{mgR/k}

❌ Check the radial equation. With

N = 4mg

the inward force is

mg + 4mg = 5mg

, giving

v_{top}^2 = 5gR

and

\tfrac{1}{2}mv_{top}^2 = \tfrac{5}{2}mgR

— the

\tfrac{3}{2}mgR

of the video belonged to

N = 2mg

❌ Check the energy budget. The spring must pay for both the climb

2mgR

and the kinetic energy

\tfrac{5}{2}mgR

, and

\tfrac{1}{2}kx^2 = \tfrac{9}{2}mgR

gives

x^2 = 9mgR/k

Show solution

Step 1 — Newton's Second Law at the top (both forces toward the center, $+y$ up):

-mg - 4mg = -\frac{mv_{top}^2}{R} \quad\Longrightarrow\quad 5mg = \frac{mv_{top}^2}{R} \quad\Longrightarrow\quad v_{top}^2 = 5gR

\tfrac{1}{2}mv_{top}^2 = \tfrac{5}{2}mgR

Step 2 — Energy conservation between the compressed spring and the top:

\tfrac{1}{2}kx^2 = 2mgR + \tfrac{1}{2}mv_{top}^2 = 2mgR + \tfrac{5}{2}mgR = \tfrac{9}{2}mgR

Step 3 — Solve for $x$ (multiply by $2/k$ , then take the square root):

x^2 = \frac{9mgR}{k} \quad\Longrightarrow\quad x = 3\sqrt{\frac{mgR}{k}}

Compare with the video's $\sqrt{7mgR/k} \approx 2.65\sqrt{mgR/k}$ : the climb term $2mgR$ never changed, only the kinetic share at the top.

Problem 4 · Doubling the Mass

Given: the same spring ( $k$ ) and loop ( $R$ ), but a block of mass $2m$ , again launched so that the normal force at the top is twice its weight — find how $v_{top}$ and the required compression $x$ change.

Speed at the top?

Required compression?

✅ Correct! Mass cancels in the radial equation but not in the energy equation:

v_{top} = \sqrt{3gR}

is untouched, while

x = \sqrt{7mgR/k}

grows like

\sqrt{m}

❌ The mass cancels here.

3(2m)g = (2m)v_{top}^2/R

divides through by

2m

to leave

v_{top}^2 = 3gR

— the same speed as before.

❌ Look at how $m$ sits in the formula.

x = \sqrt{7mgR/k}

depends on the square root of the mass, so doubling

m

multiplies

x

\sqrt{2} \approx 1.41

, not by

2

Show solution

Speed. With mass $2m$ the condition " $N$ equals twice the weight" reads $N = 2(2m)g$ , and the radial equation is

-(2m)g - 2(2m)g = -\frac{(2m)v_{top}^2}{R} \quad\Longrightarrow\quad 3g = \frac{v_{top}^2}{R}

Every term carries the same factor $2m$ , so it cancels: $v_{top} = \sqrt{3gR}$ , exactly as before. This is why part (b) of the video could answer in terms of $g$ and $R$ alone.

Compression. Energy is a different story — the spring must lift a heavier block and give it more kinetic energy:

\tfrac{1}{2}kx^2 = \tfrac{7}{2}(2m)gR \quad\Longrightarrow\quad x = \sqrt{\frac{7(2m)gR}{k}} = \sqrt{2}\,\sqrt{\frac{7mgR}{k}}

So $x$ is multiplied by $\sqrt{2} \approx 1.41$ . Both energy terms, $2mgR$ and $\tfrac{3}{2}mgR$ , are proportional to $m$ , so the whole budget doubles while $x$ only grows like its square root.

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