Irodov 1.137: Shifted Pivot Becomes a Vertical Circle

Problem Irodov 1.137

A ball of mass $m$ hangs at rest at the end of a thread of length $l$ , the other end of the thread held at a fixed point. That point of suspension is then set moving with constant horizontal velocity. Find the minimum value of this velocity for which the ball goes on to travel all the way around a full vertical circle of radius $l$ about the (now steadily moving) point of suspension, keeping the thread taut throughout. For that minimum case, find the tension in the thread at the instant the thread is horizontal.

The circle of radius l the ball describes about the suspension point O, drawn in the frame moving with O. It starts at the bottom with speed v_; at the horizontal position the tension T and weight m g act. — The circle of radius $l$ the ball describes about the suspension point $O$ , drawn in the frame moving with $O$ . It starts at the bottom with speed $\vect v_{\min}$ ; at the horizontal position the tension $\vect T$ and weight $m\vect g$ act.

Given

Find

the minimum shift velocity, and the tension when the thread is horizontal

Solution

Step 1Reducing to a vertical circle in the pivot’s own frame

The instant the point of suspension acquires its horizontal velocity, the ball is still momentarily at rest in the ground frame, so relative to the pivot it starts with a horizontal speed equal in magnitude to the shift velocity. In the pivot’s own (inertial) frame this is precisely a ball on a thread of length $l$ , launched horizontally from the lowest point of a vertical circle of radius $l$ about the fixed point $O$ , required to keep the thread taut all the way around. The shift velocity asked for is therefore the minimum such launch speed, $v_{\min}$ , shown at the bottom of the figure.

Step 2The critical condition at the top of the circle

A thread can only pull, never push, so the least demanding point of the circle is the top, where gravity and tension both point toward the center. Newton’s second law there, radially, is

T_{\mathrm{top}} + mg = \frac{mv_{\mathrm{top}}^2}{l}.

The minimum case is the thread just going slack at the top, $T_{\mathrm{top}}=0$ .

v_{\mathrm{top}}^2 = gl

(1)

Step 3Minimum launch speed from energy conservation

The thread does no work (it stays perpendicular to the velocity throughout), so mechanical energy is conserved from the bottom of the circle to the top, a rise of $2l$ .

\begin{align*}\frac{1}{2}mv_{\min}^2 &= \frac{1}{2}mv_{\mathrm{top}}^2 + mg(2l) \\ v_{\min}^2 &= v_{\mathrm{top}}^2 + 4gl\end{align*}

Substituting (1),

v_{\min}^2 = gl+4gl = 5gl.

v_{\min} = \sqrt{5gl}

(2)

Step 4Speed at the horizontal position

Energy conservation from the bottom to the horizontal position, a rise of $l$ , gives the speed there.

\begin{align*}\frac{1}{2}mv_h^2 &= \frac{1}{2}mv_{\min}^2 - mgl \\ v_h^2 &= v_{\min}^2 - 2gl\end{align*}

With $v_{\min}^2=5gl$ from (2),

v_h^2 = 5gl-2gl = 3gl.

Step 5Tension when the thread is horizontal

Newton’s second law along the (horizontal) thread direction reads

\begin{align*}T &= \frac{mv_h^2}{l} \\ &= \frac{m(3gl)}{l}\end{align*}

T = 3mg

(3)

Check

Dimensionally $\sqrt{gl}$ is a speed and $3mg$ is a force, as required. Newton’s second law along the radius at a general angle $\varphi$ measured from the top, combined with $v(\varphi)^2=gl(3-2\cos\varphi)$ from energy conservation, gives $T(\varphi)=3mg(1-\cos\varphi)$ : this correctly reproduces $T=0$ at the top ( $\varphi=0$ , Step 2’s own boundary condition) and $T=3mg$ at the horizontal position ( $\varphi=90^\circ$ , (3)), and reaches its largest value $T_{\mathrm{bot}}=6mg$ at the bottom ( $\varphi=180^\circ$ ), consistent with the bottom being the most demanding point on the circle rather than the top.

Remark

The tension is not simply $mg$ plus a centripetal term evaluated with the local weight component, because at the horizontal position gravity has no radial component at all; the entire centripetal demand $mv_h^2/l=3mg$ falls on the thread alone, while gravity’s full weight $mg$ acts purely to decelerate the ball tangentially at that instant. The two roles, radial support and tangential slowing, are cleanly separated only at the very top, bottom and sides of the circle.