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

An arrow should be drawn pointing towards the centre of the turntable to indicate the direction of the resultant force acting on the block.

Using the formula for centripetal force:

$F = m a_c$, where

a_c = \frac{v^2}{r}

Here, the linear velocity $v$ can be calculated as:

$v = \omega r = 1.8 \times 0.25 = 0.45 \text{ m/s}$.

Now, substituting into the centripetal acceleration formula:

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

Therefore, the resultant force is:

$F = 0.12 \times 0.81 = 0.0972 \text{ N}$.

According to Newton's first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by a resultant force. In this scenario, since the block is moving in a circular path, there is a change in direction, indicating that a resultant force must be acting upon it to maintain this curved motion.

The period of the pendulum is the same as the time it takes the block to complete one revolution, which is 2.5 s. The length of the pendulum can be calculated using the formula:

$T = 2\pi \sqrt{\frac{L}{g}}$, where $g \approx 9.81 \text{ m/s}^2$. Rearranging gives:

$L = \frac{g T^2}{4\pi^2}$. Substituting the values:

$L = \frac{9.81 \times (2.5)^2}{4\pi^2} \approx 0.155 \text{ m}$.

With air resistance affecting the pendulum, the amplitude of its motion may decrease over time, leading to smaller oscillations compared to the block's constant amplitude. This damping effect will cause a phase lag in the pendulum's shadow compared to the block's shadow, resulting in a loss of synchronization between the two.