Question

1.a small ball of mass m is attached to a light string of length r and whirled in a vertical circle. at the top of the circle, the ball has just enough speed so that the string remains taut. at that instant, the string breaks, and the ball flies off as a projectile. (a) derive an expression for the minimum speed vt of the ball at the top of the circle such that the string remains taut. (b) at the instant the string breaks, the center of the circle is at a height h above the ground. determine the horizontal and vertical components of the velocity of the ball just after the string breaks. (c) using your result from (a), determine the time of flight and horizontal range of the ball before it hits the ground, in terms of r, h and g. (d) qualitatively describe how the horizontal range would change if the string were to break when the ball is at the bottom of the circle instead of at the top, assuming the same circular speed at that point.

Explanation:

Step1: Analyze forces at top (for part a)

At the top of the vertical circle, the net force towards the center provides the centripetal force. The forces acting on the ball are the tension \(T\) (minimum when \(T = 0\) for the string to remain taut) and the weight \(mg\). Using the centripetal - force formula \(F_c=\frac{mv^{2}}{R}\), when \(T = 0\), \(mg=\frac{mv_{t}^{2}}{R}\).

Step2: Solve for \(v_t\) (for part a)

From \(mg=\frac{mv_{t}^{2}}{R}\), we can cancel out the mass \(m\) (since \(m
eq0\)) on both sides of the equation. Then, we get \(v_{t}^{2}=gR\), and taking the square - root of both sides, \(v_{t}=\sqrt{gR}\).

Step3: Determine velocity components (for part b)

When the string breaks at the top of the circle, the ball moves as a projectile. The velocity of the ball at the top of the circle is horizontal. So, the horizontal component of velocity \(v_x\) is equal to the speed at the top of the circle \(v_t\), and the vertical component of velocity \(v_y = 0\). So, \(v_x=\sqrt{gR}\) and \(v_y = 0\).

Step4: Use vertical - motion equation (for part c - time of flight)

The vertical displacement of the ball \(y=h - R\) (the center of the circle is at height \(h\) and the radius of the circle is \(R\)). Using the vertical - motion equation \(y=v_y t+\frac{1}{2}gt^{2}\), since \(v_y = 0\), we have \(h - R=\frac{1}{2}gt^{2}\). Solving for \(t\), we get \(t=\sqrt{\frac{2(h - R)}{g}}\).

Step5: Use horizontal - motion equation (for part c - horizontal range)

The horizontal range \(x\) is given by the formula \(x = v_x t\). We know \(v_x=\sqrt{gR}\) and \(t=\sqrt{\frac{2(h - R)}{g}}\). Substituting the values, \(x=\sqrt{gR}\times\sqrt{\frac{2(h - R)}{g}}=\sqrt{2R(h - R)}\).

Step6: Analyze bottom - of - the - circle case (for part d)

At the bottom of the circle, using the conservation of mechanical energy \(E=\frac{1}{2}mv^{2}\). The speed \(v_b\) at the bottom of the circle (from \(mg(2R)=\frac{1}{2}mv_{b}^{2}-\frac{1}{2}mv_{t}^{2}\), with \(v_t=\sqrt{gR}\)) is greater than \(v_t\) (\(v_{b}=\sqrt{5gR}\)). The time of flight \(t=\sqrt{\frac{2(h + R)}{g}}\) (because the center of the circle is at height \(h\) and the ball is at a distance \(R\) below the center at the bottom). Since the horizontal range \(x = v_x t\), and \(v_x\) (speed at the point of break) and \(t\) (time of flight) are both larger when the string breaks at the bottom (compared to the top where \(v_x=\sqrt{gR}\) and \(t=\sqrt{\frac{2(h - R)}{g}}\)), the horizontal range would increase.

Answer:

(a) \(v_{t}=\sqrt{gR}\)
(b) Horizontal component \(v_x=\sqrt{gR}\), vertical component \(v_y = 0\)
(c) Time of flight \(t=\sqrt{\frac{2(h - R)}{g}}\), horizontal range \(x=\sqrt{2R(h - R)}\)
(d) The horizontal range would increase.