A small trolley of mass 2.0 kg resting on a horizontal turntable is connected by a light spring to the centre of the table. When the turntable is set into rotation at a speed of 300 rev/min the length of the stretched spring is 40 cm. If the original length of the spring is 35 cm, determine the force constant of the spring.

Explanation

To solve the problem, we need to determine the force constant (k) of the spring when a small trolley of mass 2.0 kg is rotating on a turntable. Here are the steps to find the solution:

### Step 1: Convert the rotational speed from revolutions per minute to radians per second.
The given speed is 300 revolutions per minute (rev/min). We can convert this to radians per second (rad/s) using the following conversion factors:

\[ \text{Angular speed} (\omega) = \text{speed in rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \]

Calculating this gives:

\[ \omega = 300 \times \frac{2\pi}{60} = 10\pi \text{ rad/s} \]

### Step 2: Determine the extension of the spring.
The original length of the spring is 35 cm, and the length of the stretched spring is 40 cm. The extension (x) of the spring can be calculated as:

\[ x = \text{stretched length} - \text{original length} = 40 \text{ cm} - 35 \text{ cm} = 5 \text{ cm} = 0.05 \text{ m} \]

### Step 3: Use the centripetal force equation.
When the trolley rotates, the centripetal force required to keep it moving in a circle is provided by the spring force. The centripetal force (F_c) can be expressed as:

\[ F_c = m \cdot r \cdot \omega^2 \]

Where:
- \(m\) = mass of the trolley = 2.0 kg
- \(r\) = stretched length of the spring = 0.4 m
- \(\omega\) = angular velocity = \(10\pi \text{ rad/s}\)

Calculating the centripetal force:

\[ F_c = 2.0 \cdot 0.4 \cdot (10\pi)^2 \]

\[ F_c = 2.0 \cdot 0.4 \cdot 100\pi^2 = 0.8 \cdot 100\pi^2 = 80\pi^2 \text{ N} \]

### Step 4: Relate the spring force to the force constant.
The force exerted by the spring (F_s) is given by Hooke's Law:

\[ F_s = k \cdot x \]

Setting the centripetal force equal to the spring force:

\[ 80\pi^2 = k \cdot 0.05 \]

### Step 5: Solve for the force constant (k).
Rearranging the equation to solve for \(k\):

\[ k = \frac{80\pi^2}{0.05} \]

Calculating \(k\):

\[ k = 1600\pi^2 \text{ N/m} \]

Using \(\pi \approx 3.14\):

\[ k \approx 1600 \cdot (3.14)^2 \approx 1600 \cdot 9.86 \approx 15776 \text{ N/m} \]

### Final Answer:
The force constant of the spring is approximately \(k \approx 15776 \text{ N/m}\).

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