When a String Breaks: Which Way Does the Object Move?
Learn why an object released from circular motion initially travels along the tangent, not directly away from the centre. Worked examples and HSC-style questions explain how gravity changes its path after release.
You’re swinging a ball on a string in a circle when the string snaps. Does the ball shoot directly away from the centre, continue along the string’s old line, or head off in another direction? Make a prediction before reading on.
The ball initially travels along the tangent to the circle. That’s the straight line that touches the circle at the release point and is perpendicular to the radius. It does not launch directly away from the centre.
When the string breaks: tangent or radial?
| Circular motion before release | v ⟂ r The ball moves clockwise around a circular path. At the top, the radius points upwards from the centre to the ball. |
|---|---|
| Correct: tangent velocity | At release, the velocity arrow points right, along the tangent. The ball initially continues in its instantaneous direction of motion. |
| Incorrect: radial-outward prediction | The outward arrow points upwards, directly away from the centre. The ball does not suddenly move radially outward when tension disappears. |
01The object is already moving along the tangent
The key is to separate two things: the direction an object is moving and the direction its velocity is changing.
At every instant, a moving object has a velocity. Its direction tells you which way it is travelling at that instant. When an object moves in a circle, its velocity points along the tangent, not towards the centre.
The string points towards the centre. Its tension pulls the object inward, continually changing the direction of the object’s velocity and keeping its path curved.
Imagine a mate steering you around a chair by tugging your backpack sideways. If they suddenly let go, you don’t leap directly away from the chair. You keep moving in the direction you were already walking.
The analogy isn’t exact, because a person can stumble or turn themselves. But it captures the important idea: the sideways pull changes your direction of motion. It doesn’t create an outward launch.
Why does the velocity keep changing?
Picture the ball at two very close instants during its circular motion.
At the first instant, its velocity points along one tangent. A moment later, the velocity points along a slightly different tangent. If the ball is moving at constant speed, the two velocity vectors have the same length, but different directions.
How inward force changes velocity
| Visual 1: Ball on a circular path | Show three nearby clockwise positions around a marked centre. At each position, draw equal-length tangent velocity arrows and inward tension arrows. |
|---|---|
| Visual 2: Change in velocity | Δv = v₂ − v₁ Place two equal-length velocity vectors tail-to-tail, matching successive tangents. Draw Δv from the tip of v₁ to the tip of v₂; it points approximately towards the centre. |
The change in velocity is what produces acceleration. For uniform circular motion, this acceleration points towards the centre, even though the velocity points along the tangent.
This inward acceleration is called centripetal acceleration. Its magnitude is
\[ a_c=\frac{v^2}{r} \]
where:
- \(a_c\) is the centripetal acceleration, in metres per second squared (\(\mathrm{m\\,s^{-2}}\)).
- \(v\) is the object’s speed, in metres per second (\(\mathrm{m\\,s^{-1}}\)).
- \(r\) is the radius of the circle, in metres (m).
The corresponding net inward force is
\[ F_c=\frac{mv^2}{r} \]
where \(F_c\) is the net inward force, in newtons (N), and \(m\) is the object’s mass, in kilograms (kg).
Notice what these equations tell you. They describe the inward acceleration and force needed to maintain circular motion. They do not describe an outward force that throws the object away from the centre.
In a horizontal circle held by a string, tension supplies the required inward force. In other situations, other forces can contribute.
02What changes when the string breaks?
Immediately before the string breaks, the object has a tangential velocity and the string provides tension towards the centre.
Immediately after the break, the tension disappears. However, the object’s velocity does not suddenly vanish or point in a new direction.
The object was already moving along the tangent. It simply loses the force that was continually turning its velocity.
In the ideal case where no net force remains, Newton’s first law tells us that the object continues in a straight line at constant speed.
When the string breaks, which way does the ball go?
| Actual path: tangent | v ⟂ r Show clockwise circular motion before release, with the centre and release radius marked. Continue the path in a straight line tangent to the circle, with an arrow labelled “Actual path after release”. |
|---|---|
| Incorrect prediction: radial outward | From the release point, draw a separate dashed line directly away from the centre. Label it “Not the direction of motion”. |
There is one important qualification: this straight-line prediction assumes that no net force acts after release.
For example, a puck on a perfectly smooth horizontal table still experiences gravity and a normal force from the table. These vertical forces cancel, leaving no net horizontal force once the string breaks. The puck therefore travels in a straight line horizontally.
In that ideal situation:
- The velocity immediately after release is the same as immediately before release.
- The horizontal acceleration changes from centripetal acceleration to zero.
- The puck continues along the tangent at constant speed.
The acceleration can change suddenly when the string breaks, but the velocity does not jump suddenly.
Why does the object end up farther from the centre?
This is one reason the misconception is so tempting.
After release, the object may move farther and farther from the original centre. Surely that means it is accelerating directly outwards?
No. Increasing distance from the centre does not necessarily mean there is an outward force or outward acceleration.
The object is travelling along a straight tangent. That straight line gradually takes it farther from the centre, even though its velocity remains constant.
We’ll calculate this explicitly in Question 4.
03What if gravity or air resistance is still acting?
The ideal straight-line model is useful, but a real object usually experiences other forces after the string breaks.
The most important example is gravity. If a ball is released while moving in a vertical circle, gravity continues acting after the string breaks. Ignoring air resistance, the ball then follows a projectile trajectory rather than a straight line.
Its initial velocity is still tangent to the circle. Gravity changes the velocity after release, causing the trajectory to curve.
Gravity after release: top versus bottom
| Release at the top | v → g ↓ Mark the circle’s centre directly below the release point. Show clockwise tangential velocity to the right, gravity downward towards the centre, and a projectile path curving downwards to the right. |
|---|---|
| Release at the bottom | v ← g ↓ Mark the circle’s centre directly above the release point. Show clockwise tangential velocity to the left, gravity downward away from the centre, and a projectile path curving downwards to the left. |
Look carefully at the top and bottom of the circle.
- At the top: the velocity is horizontal, while gravity points downwards, towards the original centre. The released ball initially curves downwards towards that centre.
- At the bottom: the velocity is horizontal, while gravity again points downwards, now away from the original centre. The released ball initially curves downwards, away from that centre.
In both cases, the initial velocity is tangent to the circle. What changes is the acceleration caused by the forces that remain.
This is a subtle but important distinction: when the string breaks, tension disappears, but gravity does not. The phrase centripetal force describes the net inward force needed for circular motion; it is not a separate force that must vanish whenever the string breaks.
Air resistance can also affect the subsequent motion. It acts opposite to the object’s velocity, reducing its speed and changing its trajectory. With gravity and air resistance acting together, the path is generally more complicated than the ideal projectile parabola.
The reliable method is to identify the velocity at release, then work out the net force from that instant onwards.
04Worked example: A string snaps on a smooth table
A \(0.30\ \mathrm{kg}\) stopper moves at \(3.6\ \mathrm{m\\,s^{-1}}\) in a horizontal circle of radius \(0.90\ \mathrm{m}\) on a smooth table. The string snaps.
Calculate the centripetal acceleration and string tension immediately before the snap. Then describe the stopper’s horizontal motion immediately afterwards. Ignore air resistance.
Step 1: Calculate the centripetal acceleration before the string breaks.
Use
\[ a_c=\frac{v^2}{r} \]
Substitute the speed and radius:
\[ \begin{aligned} a_c&=\frac{(3.6)^2}{0.90}\\\ &=14.4\ \mathrm{m\\,s^{-2}} \end{aligned} \]
The stopper has a centripetal acceleration of \(14.4\ \mathrm{m\\,s^{-2}}\), directed towards the centre.
Step 2: Calculate the string tension.
Before the break, tension supplies the net horizontal centripetal force. Using \(F_c=ma_c\),
\[ \begin{aligned} T&=ma_c\\\ &=0.30\times14.4\\\ &=4.32\ \mathrm{N} \end{aligned} \]
The string tension is \(4.32\ \mathrm{N}\), directed towards the centre.
Step 3: Determine the motion after the break.
Once the string snaps, tension becomes zero. The table’s normal force balances the stopper’s weight, and friction is negligible. There is no net horizontal force, so the horizontal acceleration is zero.
The stopper continues along the tangent at its original speed of \(3.6\ \mathrm{m\\,s^{-1}}\).
The key result is that its horizontal acceleration changes from \(14.4\ \mathrm{m\\,s^{-2}}\) towards the centre to zero. Its velocity remains unchanged at the instant of release.
05Worked example: A ball released from the top of a vertical circle
A \(0.20\ \mathrm{kg}\) ball moves to the right at the top of a vertical circle of radius \(1.0\ \mathrm{m}\), with a speed of \(5.0\ \mathrm{m\\,s^{-1}}\). The string breaks at the top.
Ignoring air resistance, calculate the string tension immediately before the break, then find the ball’s displacement and velocity \(0.40\ \mathrm{s}\) after release.
Step 1: Calculate the centripetal acceleration before the break.
At the top of the circle, the centre is directly below the ball. The required centripetal acceleration is therefore downwards.
\[ \begin{aligned} a_c&=\frac{v^2}{r}\\\ &=\frac{(5.0)^2}{1.0}\\\ &=25\ \mathrm{m\\,s^{-2}} \end{aligned} \]
Step 2: Calculate the string tension.
At the top, both tension and gravity point downwards. The net inward force must equal \(mv^2/r\), so
\[ T+mg=\frac{mv^2}{r} \]
Rearrange to find the tension:
\[ \begin{aligned} T&=m\left(\frac{v^2}{r}-g\right)\\\ &=0.20(25-9.8)\\\ &=3.04\ \mathrm{N} \end{aligned} \]
The tension immediately before the break is \(3.04\ \mathrm{N}\).
Step 3: Identify the forces and initial velocity after release.
Once the string breaks, tension disappears, but gravity remains. The ball’s initial velocity is horizontal, to the right.
Choose the positive \(x\)-direction to the right and the positive \(y\)-direction upwards. The initial velocity components are
\[ v_{0x}=5.0\ \mathrm{m\\,s^{-1}}, \qquad v_{0y}=0 \]
The acceleration components after release are
\[ a_x=0,\qquad a_y=-9.8\ \mathrm{m\\,s^{-2}} \]
Step 4: Calculate the displacement after \(0.40\ \mathrm{s}\).
The horizontal displacement is
\[ \begin{aligned} x&=v_{0x}t\\\ &=5.0\times0.40\\\ &=2.0\ \mathrm{m} \end{aligned} \]
The vertical displacement is
\[ \begin{aligned} y&=v_{0y}t-\frac12gt^2\\\ &=0-\frac12(9.8)(0.40)^2\\\ &=-0.784\ \mathrm{m} \end{aligned} \]
After \(0.40\ \mathrm{s}\), the ball is \(2.0\ \mathrm{m}\) to the right and \(0.784\ \mathrm{m}\) below its release point.
Step 5: Calculate the velocity after \(0.40\ \mathrm{s}\).
The horizontal velocity remains constant:
\[ v_x=5.0\ \mathrm{m\\,s^{-1}} \]
Gravity changes the vertical component:
\[ \begin{aligned} v_y&=v_{0y}-gt\\\ &=0-9.8(0.40)\\\ &=-3.92\ \mathrm{m\\,s^{-1}} \end{aligned} \]
The speed is the magnitude of the velocity vector:
\[ \begin{aligned} v&=\sqrt{v_x^2+v_y^2}\\\ &=\sqrt{5.0^2+3.92^2}\\\ &=6.35\ \mathrm{m\\,s^{-1}} \end{aligned} \]
The angle below the horizontal is
\[ \theta=\tan^{-1}\left(\frac{3.92}{5.0}\right) =38.1^\circ \]
The ball is now travelling at \(6.35\ \mathrm{m\\,s^{-1}}\), \(38.1^\circ\) below the horizontal.
The ball did initially travel along the tangent. It did not continue along that straight line because gravity changed its velocity after release.
From circular motion to projectile motion
| Immediately before the string breaks | v = 5.0 m/s → Show the ball at the top of a vertical circle, with the centre 1.0 m directly below it and radius r pointing downwards. Draw downward tension T = 3.04 N and weight mg = 1.96 N arrows; label the net inward force 5.00 N. |
|---|---|
| 0.40 s after release | Δx = 2.0 m; Δy = 0.784 m down Draw the projectile path curving downwards from the release point, initially tangent to the circle and directed right. Mark the 0.40 s position 2.0 m right and 0.784 m below release; show gravity g = 9.8 m/s² vertically downwards. |
06Questions and solutions
Question 1
A bead moves anticlockwise around a horizontal circle, viewed from above. The string snaps when the bead is at the rightmost point of the circle.
Which direction does the bead initially travel: up the page, down the page, directly right, or directly left? Explain why.
Solution 1
The bead initially travels up the page, along the tangent to the circle.
At the rightmost point, the radius points directly right, towards the bead’s position from the centre. Because the bead is moving anticlockwise, its velocity is perpendicular to the radius and points upwards.
The tempting answer is directly right, away from the centre. But that is a radial direction, not the direction of the bead’s existing velocity. The string had been pulling the bead inwards to change its direction of motion; it had not been launching it outwards.
Question 2
A ball leaves a circular path with a speed of \(8.0\ \mathrm{m\\,s^{-1}}\). At the instant of release, its tangent is directed \(30^\circ\) above the horizontal.
Ignoring air resistance, calculate:
(a) the horizontal and vertical displacement after \(0.50\ \mathrm{s}\);
(b) the horizontal and vertical velocity components after \(0.50\ \mathrm{s}\);
(c) the ball’s speed and direction after \(0.50\ \mathrm{s}\).
Take \(g=9.8\ \mathrm{m\\,s^{-2}}\).
Solution 2
After \(0.50\ \mathrm{s}\), the ball is \(3.46\ \mathrm{m}\) horizontally to the right and \(0.775\ \mathrm{m}\) above its release point. Its speed is \(6.99\ \mathrm{m\\,s^{-1}}\), directed approximately \(7.4^\circ\) below the horizontal.
The ball starts with a velocity tangent to the circle, but gravity changes its vertical velocity throughout the flight.
(a) Displacement
Resolve the initial velocity into horizontal and vertical components:
\[ \begin{aligned} v_{0x}&=8.0\cos30^\circ\\\ &=6.93\ \mathrm{m\\,s^{-1}}\\\\[4pt] v_{0y}&=8.0\sin30^\circ\\\ &=4.00\ \mathrm{m\\,s^{-1}} \end{aligned} \]
Horizontal acceleration is zero, so
\[ \begin{aligned} x&=v_{0x}t\\\ &=6.93(0.50)\\\ &=3.46\ \mathrm{m} \end{aligned} \]
Vertically, gravity acts downwards:
\[ \begin{aligned} y&=v_{0y}t-\frac12gt^2\\\ &=4.00(0.50)-\frac12(9.8)(0.50)^2\\\ &=2.00-1.225\\\ &=0.775\ \mathrm{m} \end{aligned} \]
The positive vertical displacement means the ball is still above its release height.
(b) Velocity components
The horizontal component stays constant:
\[ v_x=6.93\ \mathrm{m\\,s^{-1}} \]
The vertical component changes under gravity:
\[ \begin{aligned} v_y&=v_{0y}-gt\\\ &=4.00-9.8(0.50)\\\ &=-0.90\ \mathrm{m\\,s^{-1}} \end{aligned} \]
The negative result means the ball is moving downwards at this instant. It has passed its highest point.
(c) Speed and direction
The speed is
\[ \begin{aligned} v&=\sqrt{v_x^2+v_y^2}\\\ &=\sqrt{6.93^2+(-0.90)^2}\\\ &=6.99\ \mathrm{m\\,s^{-1}} \end{aligned} \]
The angle below the horizontal is
\[ \theta=\tan^{-1}\left(\frac{0.90}{6.93}\right) =7.4^\circ \]
A common mistake is to assume that the ball continues along the initial tangent for the entire \(0.50\ \mathrm{s}\). The tangent gives the initial direction only. Gravity changes the vertical component of velocity, so the trajectory bends away from that straight line.
Question 3
A \(0.40\ \mathrm{kg}\) ball moves at \(4.0\ \mathrm{m\\,s^{-1}}\) at the bottom of a vertical circle of radius \(0.80\ \mathrm{m}\). The string breaks at the bottom.
(a) Calculate the net inward force and the string tension immediately before the break.
(b) State the ball’s acceleration immediately after the break, relative to the original centre of the circle.
(c) Explain why the direction of this acceleration does not determine the ball’s initial direction of travel.
Ignore air resistance and use \(g=9.8\ \mathrm{m\\,s^{-2}}\).
Solution 3
The net inward force immediately before the break is \(8.0\ \mathrm{N}\) upwards, and the tension is \(11.92\ \mathrm{N}\) upwards. Immediately after the break, the ball accelerates downwards, away from the original centre, but its initial velocity remains horizontal and tangent to the circle.
(a) Net inward force and tension
At the bottom of the circle, the centre is above the ball. The required centripetal acceleration therefore points upwards.
\[ \begin{aligned} a_c&=\frac{v^2}{r}\\\ &=\frac{(4.0)^2}{0.80}\\\ &=20\ \mathrm{m\\,s^{-2}} \end{aligned} \]
The net inward force is
\[ \begin{aligned} F_c&=ma_c\\\ &=0.40\times20\\\ &=8.0\ \mathrm{N} \end{aligned} \]
Tension acts upwards, while the ball’s weight acts downwards. Therefore,
\[ T-mg=F_c \]
Rearranging,
\[ \begin{aligned} T&=F_c+mg\\\ &=8.0+0.40(9.8)\\\ &=11.92\ \mathrm{N} \end{aligned} \]
The tension must be greater than the net inward force because it must also balance the downward weight and still provide a net force upwards.
(b) Acceleration after the break
Once the string breaks, tension disappears. Gravity is the only force acting, so the acceleration is \(9.8\ \mathrm{m\\,s^{-2}}\) downwards.
Since the centre of the original circle is above the ball, this acceleration points away from that centre.
(c) Acceleration is not velocity
The ball’s velocity at the bottom of the circle is horizontal because it is tangent to the circle there. Its acceleration after release is vertical and downwards.
Acceleration describes how velocity changes, not the direction in which the object must already be moving. The ball initially moves horizontally, then curves downwards under gravity.
The tempting mistake is to interpret an outward acceleration as an immediate outward velocity. These are different vectors.
Question 4
A puck moves at \(3.0\ \mathrm{m\\,s^{-1}}\) in a circle of radius \(0.80\ \mathrm{m}\) on a frictionless horizontal table. The string breaks, and the puck travels along the tangent at constant speed.
How far from the original centre of the circle is the puck after \(0.60\ \mathrm{s}\)? Does the increase in distance from the centre mean that the puck has an outward acceleration? Explain.
Solution 4
The puck is approximately \(1.97\ \mathrm{m}\) from the original centre after \(0.60\ \mathrm{s}\). No, this does not mean it has an outward acceleration. Its horizontal acceleration is zero; the increase in distance is a consequence of its straight-line path.
The puck travels along the tangent, which is perpendicular to the radius at the release point.
Why the puck moves farther from the centre
| Tangent displacement forms a right-angled triangle | d = √(r² + s²) ≈ 1.97 m Draw circle centre O, release point P, radius OP = 0.80 m, and tangent displacement PQ = 1.80 m. Mark the right angle at P and join O to Q, labelling OQ = d ≈ 1.97 m. |
|---|
Step 1: Calculate the distance travelled along the tangent.
\[ \begin{aligned} s&=vt\\\ &=3.0(0.60)\\\ &=1.80\ \mathrm{m} \end{aligned} \]
Step 2: Use the right-angled triangle.
The radius and the tangent displacement are perpendicular. The distance \(d\) from the original centre to the puck’s later position is therefore found using Pythagoras’ theorem:
\[ \begin{aligned} d&=\sqrt{r^2+s^2}\\\ &=\sqrt{(0.80)^2+(1.80)^2}\\\ &=\sqrt{3.88}\\\ &=1.97\ \mathrm{m} \end{aligned} \]
The puck has moved from a distance of \(0.80\ \mathrm{m}\) from the centre to \(1.97\ \mathrm{m}\).
Step 3: Interpret the result.
The puck’s distance from the centre increases, but its velocity vector remains constant. There is no net horizontal force, so there is no horizontal acceleration.
The crucial point is that distance from the centre is not the same thing as displacement along the path. A straight tangent can carry an object farther from the centre without any outward force acting on it.
Question 5
A ball is moving clockwise at the top of a vertical circle of radius \(1.00\ \mathrm{m}\). Its speed is
\[ v=\sqrt{gr}=3.13\ \mathrm{m\\,s^{-1}} \]
The string is cut exactly at the top. Ignore air resistance.
Compare two possible motions:
- Case A: the string remains intact, so the ball continues around the circle.
- Case B: the string is cut, so the ball moves freely under gravity.
A student argues that the ball in Case B must continue following the same circle for at least a short time because gravity provides exactly the centripetal acceleration needed at the top.
Is this conclusion correct? Explain what is identical at the instant of release and why the two trajectories subsequently diverge.
Solution 5
No. The two motions have the same position, velocity, and acceleration at the instant the string is cut, but they do not follow the same path after that instant.
This is a more subtle situation than simply removing the inward force. At the top of the circle, gravity itself supplies the required centripetal acceleration.
Step 1: Compare the required acceleration with gravity.
The centripetal acceleration required at the top is
\[ a_c=\frac{v^2}{r}=g \]
Its direction is downwards, towards the centre of the circle. Gravity also acts downwards, with acceleration \(g\).
In Case A, the net inward force must satisfy
\[ T+mg=\frac{mv^2}{r} \]
Since \(v^2/r=g\),
\[ T+mg=mg \]
Therefore,
\[ T=0 \]
At the exact top of the circle, the string tension is zero. Gravity alone supplies the required centripetal acceleration.
In Case B, once the string is cut, gravity is also the only force acting. The acceleration is again \(g\) downwards.
The student’s reasoning is tempting because the two accelerations really are equal at the instant of release.
Step 2: Work out what happens after that instant.
Same starting motion, different trajectories
| String intact: circular motion | a_c = v²/r Start at the top with a rightward tangent velocity and downward inward acceleration; initial tension T = 0. Draw the circular path curving down and right. At a later point, show inward acceleration directed down and left. |
|---|---|
| String cut: projectile motion | a = g ↓ Start at the same point with the same rightward velocity and downward acceleration. Draw a projectile path curving down and right, separating from the circle; gravity remains vertically downward. |
In Case A, if the string remains intact, the ball continues around the circle. As it moves slightly to the right of the top, the direction towards the centre tilts downwards and to the left. The inward acceleration required to maintain the circle must turn with it, and the string develops tension to help provide that force.
In Case B, there is no string tension. Gravity continues acting vertically downwards, regardless of the ball’s position. It cannot supply the changing inward force needed to keep the ball on the original circular path.
The free ball therefore follows a projectile trajectory. It starts at the same position, with the same velocity and the same instantaneous acceleration as the ball in Case A, but the paths diverge after release.
The mistake is assuming that matching the acceleration at one instant guarantees the same motion afterwards. Circular motion requires the direction of the net acceleration to change continuously. Gravity alone cannot maintain that changing direction.
This is the useful link to projectile motion: circular motion tells you the velocity at the instant of release, while Newton’s laws and the projectile equations tell you what happens next.