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Step-by-Step Solution
Step 1: Understand the Physical Situation
A 200 g block moves along a horizontal circular groove of radius 20 cm. It completes one revolution in 40 s, moving with uniform speed. We want to find the normal force by the side walls on the block.
Step 2: Identify the Force Providing Centripetal Acceleration
Since the block is moving in a circular path, it requires a centripetal force directed towards the center of the circular groove. This force is provided by the normal force (N) from the vertical walls of the groove.
Step 3: Write the Expression for Centripetal Force
The necessary centripetal force on the block of mass $m$ moving with angular speed $\omega$ in a circle of radius $R$ is given by:
$ F_{c} \;=\; m \omega^{2} R. $
Here, $F_{c}$ is provided entirely by the normal force. Hence:
$ N \;=\; m \omega^{2} R. $
Step 4: Compute the Angular Velocity
The time period $T$ for one revolution is 40 s. The angular velocity $\omega$ is:
$$
\omega \;=\; \frac{2\pi}{T}.
$$
Step 5: Substitute Known Values
• Mass, $m = 0.2 \text{ kg}$ (since 200 g = 0.2 kg)
• Radius of circular path, $R = 0.2 \text{ m}$ (since 20 cm = 0.2 m)
• Time period, $T = 40 \text{ s}$.
First, calculate $\omega$:
$$
\omega \;=\; \frac{2\pi}{40} \;=\; \frac{\pi}{20}.
$$
Then,
$$
\omega^{2}
\;=\; \left(\frac{\pi}{20}\right)^2
\;=\; \frac{\pi^2}{400}.
$$
Thus, the normal force $N$ becomes:
$$
N
\;=\; m \,\omega^{2}\, R
\;=\; 0.2
\times \frac{\pi^2}{400}
\times 0.2.
$$
Step 6: Calculate the Numerical Value
Let us take $\pi \approx 3.14$. Then:
$$
\omega^{2}
\;\approx\; \frac{(3.14)^2}{400}
\;=\; \frac{9.8596}{400}
\;=\; 0.024649.
$$
And so,
$$
N
= 0.2 \times 0.024649 \times 0.2
= 0.2 \times 0.2 \times 0.024649
\approx 9.859 \times 10^{-4} \,\text{N}.
$$
Final Answer
The normal force exerted by the walls of the groove on the block is approximately
$9.859 \times 10^{-4}\,\text{N}$.
