A teacher sets up a demonstration to show the relationship between circular motion and simple harmonic motion (SHM) - AQA - A-Level Physics - Question 5 - 2022 - Paper 1

The arrow should be drawn pointing towards the center of the turntable, indicating the direction of the resultant centripetal force acting on the block.

The resultant force can be calculated using the formula for centripetal force:

$F = m a_c$

where $m = 0.12$ kg is the mass of the block and $a_c = \frac{v^2}{r}$ is the centripetal acceleration. To find $v$, we can use $v = r \omega = 0.25 \times 1.8 = 0.45$ m/s. Thus:

$a_c = \frac{(0.45)^2}{0.25} = 0.81 \text{ m/s}^2$

Now calculating the force:

$F = 0.12 \times 0.81 \approx 0.097 \text{ N}$

According to Newton's first law of motion, an object at rest will stay at rest and an object in motion will stay in motion unless acted upon by a resultant force. In this case, the block moves in a circular path, indicating a constant change in direction, which can only occur if a resultant force is acting upon it to keep it in circular motion.

For a pendulum to complete one full oscillation in the same time period as the rotating block (2.5 s), we use the formula for the period of a simple pendulum:

$T = 2\pi \sqrt{\frac{L}{g}}$

Rearranging this gives:

$L = \frac{g T^2}{4\pi^2}$

Substituting $g = 9.81$ m/s² and $T = 2.5$ s:

$L = \frac{9.81 \times (2.5)^2}{4\pi^2} \approx 0.84 ext{ m}$

The air resistance will dampen the motion of the pendulum, resulting in reduced amplitude over time. This change may lead to a phase difference between the shadows since the block continues its motion uniformly, while the pendulum’s motion becomes slower and less pronounced.