P0741 Mechanics › Kinematics

Projectile Motion from a Circular Path

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Kinematics Intermediate relative motion

Source: Physics Competition - Mechanics

Problem Sets:

Relative motion - point mass

Problem

An ice skater skates on a circular path of radius $R = 15$ m at a constant speed $v_0 = 7$ m/s. They throw a ball horizontally from a height $h = 1.5$ m, aiming for it to land at the center of the circle.

  1. What is the required magnitude of the ball's velocity relative to the skater?
  2. What is the direction of this velocity, expressed as the angle $\theta$ with the skater's instantaneous velocity?

[Q1] $v_{bs} = 28.0$ m/s [Q2] $\theta = 104.5^\circ$

This problem combines projectile motion and relative velocity. Let $\vec{v}_s$ be the skater's velocity, $\vec{v}_{bg}$ the ball's velocity relative to the ground, and $\vec{v}_{bs}$ the ball's velocity relative to the skater.

  1. Projectile Motion: The time of flight $t$ for an object dropped from height $h$ is found from vertical motion:

    $$h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$$

    To reach the center, the ball must travel a horizontal distance $R$. The required horizontal velocity of the ball relative to the ground, $v_{bg}$, is:

    $$v_{bg} = \frac{R}{t} = R\sqrt{\frac{g}{2h}}$$

  2. Relative Velocity: The relationship between the velocities is $\vec{v}_{bg} = \vec{v}_{bs} + \vec{v}_s$, so $\vec{v}_{bs} = \vec{v}_{bg} - \vec{v}_s$. The skater's velocity $\vec{v}_s$ is tangential to the circle. The ball's required ground velocity $\vec{v}_{bg}$ is directed radially inward. Thus, $\vec{v}_s \perp \vec{v}_{bg}$. The magnitude of the relative velocity $\vec{v}_{bs}$ is found using the Pythagorean theorem:

    $$v_{bs} = \sqrt{v_{bg}^2 + (-v_s)^2} = \sqrt{\left(R\sqrt{\frac{g}{2h}}\right)^2 + v_0^2}$$

    Substituting values (with $g = 9.8$ m/s$^2$):

    $v_{bg} = 15 \sqrt{\frac{9.8}{2 \times 1.5}} \approx 27.11$ m/s. $v_{bs} = \sqrt{(27.11)^2 + (7)^2} \approx 28.0$ m/s.

    The angle $\theta$ between $\vec{v}_{bs}$ and $\vec{v}_s$ is found from the dot product or trigonometry. Since $\vec{v}_{bs} = \vec{v}_{bg} - \vec{v}_s$ and $\vec{v}_{bg} \perp \vec{v}_s$:

    $$\cos\theta = \frac{\vec{v}_{bs} \cdot \vec{v}_s}{v_{bs}v_s} = \frac{(\vec{v}_{bg} - \vec{v}_s) \cdot \vec{v}_s}{v_{bs}v_s} = \frac{-v_s^2}{v_{bs}v_s} = -\frac{v_s}{v_{bs}}$$ $\cos\theta = -7 / 28.0 = -0.25$. $\theta = \arccos(-0.25) \approx 104.5^\circ$.