Questions: A small lead ball, attached to a 1.00-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.2 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

Solution Steps

The ball is released when it is moving upward. To find the initial velocity, we need to calculate the tangential speed of the ball as it moves in the circle. The ball completes three revolutions per second, so the angular velocity \(\omega\) is:

\[ \omega = 3 \times 2\pi = 6\pi \, \text{rad/s} \]

The tangential speed \(v\) is given by:

\[ v = \omega \times r = 6\pi \times 1.00 = 6\pi \, \text{m/s} \]

The ball is released at a height of 1.2 m above the ground with an initial upward velocity of \(6\pi \, \text{m/s}\). We use the kinematic equation to find the maximum height:

\[ v_f^2 = v_i^2 + 2a(y_f - y_i) \]

At the maximum height, the final velocity \(v_f = 0\). The acceleration \(a\) is due to gravity, so \(a = -9.81 \, \text{m/s}^2\). The initial height \(y_i = 1.2 \, \text{m}\), and we need to find \(y_f\).

\[ 0 = (6\pi)^2 + 2(-9.81)(y_f - 1.2) \]

Solving for \(y_f\):

\[ 0 = 36\pi^2 - 19.62(y_f - 1.2) \]

\[ 19.62(y_f - 1.2) = 36\pi^2 \]

\[ y_f - 1.2 = \frac{36\pi^2}{19.62} \]

\[ y_f = \frac{36\pi^2}{19.62} + 1.2 \]

Calculating the value:

\[ y_f \approx \frac{36 \times 9.8696}{19.62} + 1.2 \approx 18.144 + 1.2 = 19.344 \, \text{m} \]

Final Answer