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Question: A string of length 2l is tied at one end to a fixed-point A on a frictionless horizontal surface. A ...

A string of length 2l is tied at one end to a fixed-point A on a frictionless horizontal surface. A small disc of mass m attached to the other end of the string is placed at point B as shown in the figure. The disc is given a sharp impulse, as result it starts moving with a velocity v0v_0 in a perpendicular direction to line AB. Consider the thread to be perfectly inelastic and almost inextensible.

(a) Determine the velocity of the disc after the string becomes taut.

(b) Determine the impulse of the tension force developed in the string.

(c) What would be your answers if the thread were perfectly elastic and almost inextensible.

Answer

(a) The velocity of the disc after the string becomes taut (inelastic string) has a magnitude of v02\frac{v_0}{2} and its direction is perpendicular to the string, pointing tangentially to the circular path it will subsequently follow. In vector form, v⃗after=−34v0i^+14v0j^\vec{v}_{after} = -\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j}.

(b) The impulse of the tension force developed in the string (inelastic string) has a magnitude of m32v0m \frac{\sqrt{3}}{2} v_0. In vector form, J⃗=−34mv0i^−34mv0j^\vec{J} = -\frac{\sqrt{3}}{4} m v_0 \hat{i} - \frac{3}{4} m v_0 \hat{j}.

(c) If the thread were perfectly elastic and almost inextensible: (c-a) The velocity of the disc after the string becomes taut would have a magnitude of v0v_0. In vector form, v⃗after′=−32v0i^−12v0j^\vec{v}_{after}' = -\frac{\sqrt{3}}{2} v_0 \hat{i} - \frac{1}{2} v_0 \hat{j}. (c-b) The impulse of the tension force developed in the string would have a magnitude of m3v0m \sqrt{3} v_0. In vector form, J⃗′=−32mv0i^−32mv0j^\vec{J}' = -\frac{\sqrt{3}}{2} m v_0 \hat{i} - \frac{3}{2} m v_0 \hat{j}.

Explanation

Solution

The problem describes a disc of mass mm attached to a string of length 2l2l. One end of the string is fixed at point A on a frictionless horizontal surface. The disc is initially placed at point B, such that the distance AB is ll. The disc is given a sharp impulse, resulting in an initial velocity v0v_0 perpendicular to the line AB.

1. Initial State before the string becomes taut: Let point A be the origin (0,0). Since the distance AB is ll, let B be at (l,0)(l, 0). The initial velocity of the disc is v⃗initial=v0j^\vec{v}_{initial} = v_0 \hat{j} (perpendicular to AB). Since the surface is frictionless and no other forces act until the string becomes taut, the disc moves with constant velocity v0j^v_0 \hat{j}. The position of the disc at time tt is (x(t),y(t))=(l,v0t)(x(t), y(t)) = (l, v_0 t). The string becomes taut when the distance from A to the disc equals the string length, 2l2l. So, x(t)2+y(t)2=2l\sqrt{x(t)^2 + y(t)^2} = 2l. l2+(v0t)2=2l\sqrt{l^2 + (v_0 t)^2} = 2l. Squaring both sides: l2+(v0t)2=4l2l^2 + (v_0 t)^2 = 4l^2. (v0t)2=3l2(v_0 t)^2 = 3l^2. v0t=3lv_0 t = \sqrt{3}l. The time taken for the string to become taut is t0=3lv0t_0 = \frac{\sqrt{3}l}{v_0}. At this instant, the position of the disc is (l,3l)(l, \sqrt{3}l). The velocity of the disc just before the string becomes taut is v⃗before=v0j^\vec{v}_{before} = v_0 \hat{j}.

Let's define the direction of the string at this moment. The string connects A(0,0) to the disc at (l,3l)(l, \sqrt{3}l). The vector along the string is r⃗=li^+3lj^\vec{r} = l \hat{i} + \sqrt{3}l \hat{j}. The unit vector along the string is r^=li^+3lj^2l=12i^+32j^\hat{r} = \frac{l \hat{i} + \sqrt{3}l \hat{j}}{2l} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}. Let θ\theta be the angle the string makes with the x-axis. cos⁡θ=1/2\cos \theta = 1/2, sin⁡θ=3/2\sin \theta = \sqrt{3}/2, so θ=60∘\theta = 60^\circ.

The velocity v⃗before\vec{v}_{before} can be decomposed into components along and perpendicular to the string. Component along the string (radial): vradial,before=v⃗before⋅r^=(v0j^)⋅(12i^+32j^)=32v0v_{radial, before} = \vec{v}_{before} \cdot \hat{r} = (v_0 \hat{j}) \cdot (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = \frac{\sqrt{3}}{2} v_0. Component perpendicular to the string (tangential): Let t^\hat{t} be the unit vector perpendicular to r^\hat{r} in the direction of motion. t^=−sin⁡θi^+cos⁡θj^=−32i^+12j^\hat{t} = -\sin \theta \hat{i} + \cos \theta \hat{j} = -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}. vtangential,before=v⃗before⋅t^=(v0j^)⋅(−32i^+12j^)=12v0v_{tangential, before} = \vec{v}_{before} \cdot \hat{t} = (v_0 \hat{j}) \cdot (-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) = \frac{1}{2} v_0. The initial velocity can be written as v⃗before=vradial,beforer^+vtangential,beforet^\vec{v}_{before} = v_{radial, before} \hat{r} + v_{tangential, before} \hat{t}.

(a) Determine the velocity of the disc after the string becomes taut (perfectly inelastic and almost inextensible). When the string becomes perfectly inelastic and inextensible, the component of velocity along the string (radial component) is instantaneously reduced to zero. This is because the string cannot stretch and does not allow rebound. The component of velocity perpendicular to the string (tangential component) is unaffected by the impulsive tension, as the tension acts only along the string. So, after the string becomes taut: vradial,after=0v_{radial, after} = 0. vtangential,after=vtangential,before=12v0v_{tangential, after} = v_{tangential, before} = \frac{1}{2} v_0. The velocity of the disc after the string becomes taut is v⃗after=vtangential,aftert^\vec{v}_{after} = v_{tangential, after} \hat{t}. The magnitude of the velocity is vafter=v02v_{after} = \frac{v_0}{2}. The direction of the velocity is perpendicular to the string, given by t^=−32i^+12j^\hat{t} = -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}. So, v⃗after=v02(−32i^+12j^)=−34v0i^+14v0j^\vec{v}_{after} = \frac{v_0}{2} (-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) = -\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j}.

(b) Determine the impulse of the tension force developed in the string. Impulse J⃗=Δp⃗=mv⃗after−mv⃗before\vec{J} = \Delta \vec{p} = m \vec{v}_{after} - m \vec{v}_{before}. v⃗before=v0j^\vec{v}_{before} = v_0 \hat{j}. v⃗after=−34v0i^+14v0j^\vec{v}_{after} = -\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j}. J⃗=m(−34v0i^+14v0j^−v0j^)\vec{J} = m \left( -\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j} - v_0 \hat{j} \right) J⃗=m(−34v0i^−34v0j^)\vec{J} = m \left( -\frac{\sqrt{3}}{4} v_0 \hat{i} - \frac{3}{4} v_0 \hat{j} \right) J⃗=−mv04(3i^+3j^)\vec{J} = -\frac{m v_0}{4} (\sqrt{3} \hat{i} + 3 \hat{j}).

Alternatively, since the impulse acts only along the string, it only changes the radial component of momentum. Jradial=mvradial,after−mvradial,before=m(0)−m(32v0)=−m32v0J_{radial} = m v_{radial, after} - m v_{radial, before} = m(0) - m(\frac{\sqrt{3}}{2} v_0) = -m \frac{\sqrt{3}}{2} v_0. The negative sign indicates the impulse is in the direction opposite to r^\hat{r}, i.e., towards A. The magnitude of the impulse is J=∣−m32v0∣=m32v0J = \left| -m \frac{\sqrt{3}}{2} v_0 \right| = m \frac{\sqrt{3}}{2} v_0. The impulse vector is J⃗=J(−r^)=m32v0(−12i^−32j^)=−34mv0i^−34mv0j^\vec{J} = J (-\hat{r}) = m \frac{\sqrt{3}}{2} v_0 \left( -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) = -\frac{\sqrt{3}}{4} m v_0 \hat{i} - \frac{3}{4} m v_0 \hat{j}. Both methods yield the same result.

(c) What would be your answers if the thread were perfectly elastic and almost inextensible? If the string were perfectly elastic, it would behave like an elastic collision along the line of impact (the string). In an elastic collision, the relative velocity along the line of impact reverses its direction but keeps its magnitude. Let vradial,after′v_{radial, after}' be the radial velocity component after the elastic string becomes taut. The fixed point A has zero velocity. So, the relative velocity of the disc with respect to A along the string is vradialv_{radial}. For an elastic string, the relative velocity along the string reverses: vradial,after′−vA,radial=−(vradial,before−vA,radial)v_{radial, after}' - v_{A, radial} = -(v_{radial, before} - v_{A, radial}). Since vA,radial=0v_{A, radial} = 0: vradial,after′=−vradial,before=−32v0v_{radial, after}' = -v_{radial, before} = -\frac{\sqrt{3}}{2} v_0. The tangential component of velocity remains unchanged, as it is perpendicular to the impulsive force: vtangential,after′=vtangential,before=12v0v_{tangential, after}' = v_{tangential, before} = \frac{1}{2} v_0.

(c-a) Velocity of the disc after the string becomes taut (elastic string): The velocity vector after the string becomes taut would be: v⃗after′=vradial,after′r^+vtangential,after′t^\vec{v}_{after}' = v_{radial, after}' \hat{r} + v_{tangential, after}' \hat{t} v⃗after′=(−32v0)r^+(12v0)t^\vec{v}_{after}' = (-\frac{\sqrt{3}}{2} v_0) \hat{r} + (\frac{1}{2} v_0) \hat{t} Substitute r^=12i^+32j^\hat{r} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} and t^=−32i^+12j^\hat{t} = -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}: v⃗after′=(−32v0)(12i^+32j^)+(12v0)(−32i^+12j^)\vec{v}_{after}' = (-\frac{\sqrt{3}}{2} v_0) (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) + (\frac{1}{2} v_0) (-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) v⃗after′=(−34v0i^−34v0j^)+(−34v0i^+14v0j^)\vec{v}_{after}' = (-\frac{\sqrt{3}}{4} v_0 \hat{i} - \frac{3}{4} v_0 \hat{j}) + (-\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j}) v⃗after′=(−34v0−34v0)i^+(−34v0+14v0)j^\vec{v}_{after}' = (-\frac{\sqrt{3}}{4} v_0 - \frac{\sqrt{3}}{4} v_0) \hat{i} + (-\frac{3}{4} v_0 + \frac{1}{4} v_0) \hat{j} v⃗after′=−234v0i^−24v0j^\vec{v}_{after}' = -\frac{2\sqrt{3}}{4} v_0 \hat{i} - \frac{2}{4} v_0 \hat{j} v⃗after′=−32v0i^−12v0j^\vec{v}_{after}' = -\frac{\sqrt{3}}{2} v_0 \hat{i} - \frac{1}{2} v_0 \hat{j}. The magnitude of the velocity is ∣v⃗after′∣=(−32v0)2+(−12v0)2=34v02+14v02=v02=v0|\vec{v}_{after}'| = \sqrt{(-\frac{\sqrt{3}}{2} v_0)^2 + (-\frac{1}{2} v_0)^2} = \sqrt{\frac{3}{4} v_0^2 + \frac{1}{4} v_0^2} = \sqrt{v_0^2} = v_0. So, the speed remains v0v_0. This is expected for an elastic collision with a fixed object.

(c-b) Impulse of the tension force developed in the string (elastic string): Impulse J⃗′=mv⃗after′−mv⃗before\vec{J}' = m \vec{v}_{after}' - m \vec{v}_{before}. J⃗′=m(−32v0i^−12v0j^−v0j^)\vec{J}' = m \left( -\frac{\sqrt{3}}{2} v_0 \hat{i} - \frac{1}{2} v_0 \hat{j} - v_0 \hat{j} \right) J⃗′=m(−32v0i^−32v0j^)\vec{J}' = m \left( -\frac{\sqrt{3}}{2} v_0 \hat{i} - \frac{3}{2} v_0 \hat{j} \right) J⃗′=−mv02(3i^+3j^)\vec{J}' = -\frac{m v_0}{2} (\sqrt{3} \hat{i} + 3 \hat{j}).

Alternatively, using the radial component of momentum change: Jradial′=mvradial,after′−mvradial,before=m(−32v0)−m(32v0)=−2m32v0=−m3v0J_{radial}' = m v_{radial, after}' - m v_{radial, before} = m(-\frac{\sqrt{3}}{2} v_0) - m(\frac{\sqrt{3}}{2} v_0) = -2m \frac{\sqrt{3}}{2} v_0 = -m \sqrt{3} v_0. The magnitude of the impulse is J′=m3v0J' = m \sqrt{3} v_0. The impulse vector is J⃗′=J′(−r^)=m3v0(−12i^−32j^)=−32mv0i^−32mv0j^\vec{J}' = J' (-\hat{r}) = m \sqrt{3} v_0 \left( -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) = -\frac{\sqrt{3}}{2} m v_0 \hat{i} - \frac{3}{2} m v_0 \hat{j}. Both methods yield the same result.

Summary of Answers:

(a) Velocity of the disc after the string becomes taut (inelastic string): The speed of the disc is vafter=v02v_{after} = \frac{v_0}{2}. The velocity vector is v⃗after=−34v0i^+14v0j^\vec{v}_{after} = -\frac{\sqrt{3}}{4} v_0 \hat{i} + \frac{1}{4} v_0 \hat{j}.

(b) Impulse of the tension force developed in the string (inelastic string): The magnitude of the impulse is J=m32v0J = m \frac{\sqrt{3}}{2} v_0. The impulse vector is J⃗=−34mv0i^−34mv0j^\vec{J} = -\frac{\sqrt{3}}{4} m v_0 \hat{i} - \frac{3}{4} m v_0 \hat{j}.

(c) Answers if the thread were perfectly elastic and almost inextensible: (c-a) Velocity of the disc after the string becomes taut (elastic string): The speed of the disc is vafter′=v0v_{after}' = v_0. The velocity vector is v⃗after′=−32v0i^−12v0j^\vec{v}_{after}' = -\frac{\sqrt{3}}{2} v_0 \hat{i} - \frac{1}{2} v_0 \hat{j}.

(c-b) Impulse of the tension force developed in the string (elastic string): The magnitude of the impulse is J′=m3v0J' = m \sqrt{3} v_0. The impulse vector is J⃗′=−32mv0i^−32mv0j^\vec{J}' = -\frac{\sqrt{3}}{2} m v_0 \hat{i} - \frac{3}{2} m v_0 \hat{j}.

Explanation of the solution:

  1. Determine the point and time of impact: The disc moves freely until the string becomes taut. This occurs when the distance from the fixed point A equals the string length (2l2l). Using the initial conditions, the position where the string becomes taut is determined.
  2. Decompose velocity: At the instant the string becomes taut, the disc's velocity is decomposed into two components: one along the string (radial) and one perpendicular to the string (tangential).
  3. Apply impulse-momentum principle for inelastic string:
    • For a perfectly inelastic and inextensible string, the radial component of velocity becomes zero immediately after the string becomes taut. This is because the string prevents any outward motion and does not allow rebound.
    • The tangential component of velocity remains unchanged, as the impulsive tension acts only along the string.
    • The final velocity is then composed solely of the tangential component.
    • The impulse is calculated as the change in momentum (Δp⃗=mv⃗final−mv⃗initial\Delta \vec{p} = m\vec{v}_{final} - m\vec{v}_{initial}). Since only the radial momentum changes, the impulse is purely radial.
  4. Apply impulse-momentum principle for elastic string:
    • For a perfectly elastic and inextensible string, the radial component of velocity reverses its direction while maintaining its magnitude (similar to an elastic collision with a fixed wall).
    • The tangential component of velocity remains unchanged.
    • The final velocity is then the vector sum of the reversed radial component and the unchanged tangential component.
    • The impulse is again calculated as the change in momentum.