Projectiles and Circular Motion: Quick Revision

A brief note on Projectiles and Circular Motion. Try each question at the end before you open its answer.

The big ideas

A projectile. After the launch, gravity is the only force on it (we ignore the air).

• Sideways nothing pushes, so $v_x = u\cos\theta$ never changes.
• Up and down it is free fall: $v_y = u\sin\theta - gt$.
• The path is a parabola. At the top $v_y = 0$, but $v_x$ is still there.

Uniform circular motion. A body goes round a circle at a steady speed. Its direction keeps turning, so it is accelerating. The acceleration points to the centre of the circle.

The formulas at a glance

For a launch at the speed $u$ and the angle $\theta$, over level ground:

And the path:

For a circle of radius $r$, gone round once in the time $T$:

A worked throw

A ball is kicked at $21\ \text{m/s}$, at $32^\circ$ above level ground. Take $g = 9.8\ \text{m/s}^2$.

• Parts. $u_x = 21\ \text{m/s}\cos 32^\circ = 17.81\ \text{m/s}$ and $u_y = 21\ \text{m/s}\sin 32^\circ = 11.13\ \text{m/s}$.
• Time of flight. $T = \dfrac{2u_y}{g} = \dfrac{2 \times 11.13\ \text{m/s}}{9.8\ \text{m/s}^2} = 2.27\ \text{s}$.
• Greatest height. $H = \dfrac{u_y^2}{2g} = \dfrac{(11.13\ \text{m/s})^2}{19.6\ \text{m/s}^2} = 6.32\ \text{m}$.
• Range. $R = u_x T = 17.81\ \text{m/s} \times 2.27\ \text{s} = 40.4\ \text{m}$.

Kicked at $58^\circ$ instead, it lands at the same spot, since $32^\circ + 58^\circ = 90^\circ$. It just goes higher and stays up longer.

A worked circle

A giant wheel at a fair has a radius of $9.0\ \text{m}$ and turns once every $30\ \text{s}$.

• $\omega = \dfrac{2\pi}{30\ \text{s}} = 0.209\ \text{rad/s}$.
• A rider's speed is $v = \omega r = 0.209\ \text{rad/s} \times 9.0\ \text{m} = 1.88\ \text{m/s}$.
• The rider's acceleration is $a_c = \omega^2 r = 0.395\ \text{m/s}^2$, towards the centre of the wheel.

Practice with new numbers

A program to try

This program follows the kicked ball above, every half second.

Mistakes to avoid

• "The speed at the top is zero." Only $v_y$ is zero there. The speed at the top is $u\cos\theta$.
• "The acceleration at the top is zero." It is $g$, downwards, all the way.
• $2\sin\theta$ in place of $\sin 2\theta$. The range has the sine of twice the angle.
• Degrees into Python. math.sin works in radians. Use math.radians(angle) first.
• Revolutions per minute used as radians per second. Multiply by $2\pi$ and divide by $60$ first.
• The acceleration along the circle. In uniform circular motion the velocity is along the circle, and the acceleration points to the centre, at right angles to it.

Check yourself

1. Two balls are kicked at the same speed, at $28^\circ$ and at $62^\circ$. Which lands farther? Which stays in the air longer?

2. At the very top of its path, what are a projectile's velocity and acceleration?

3. A stone is thrown horizontally at $8\ \text{m/s}$ from a bridge $19.6\ \text{m}$ above the water. How long does it take to reach the water, and how far from the point below the bridge does it land?

4. A body goes round a circle of radius $3\ \text{m}$ at a steady $6\ \text{m/s}$. Find its acceleration and the time for one turn.

5. The speed of a body in uniform circular motion doubles, and the radius stays the same. What happens to its acceleration?

6. At what launch angle is the range exactly four times the greatest height?