Question

The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30° in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be: 

• A. 100 N
• B. 9.87 N
• C. 50 N
• D. 0.49 N

Hint:

The projection of a uniform circular motion on any of its diameters is SHM. The angular velocity ($\omega$) of the circular motion is numerically equal to the angular frequency of the SHM. The maximum acceleration in SHM is always $\omega^2 A$.

Answer 1

The Correct Option is B

The restoration force per unit mass is, by definition, the acceleration ($F/m = a$).
The projection of uniform circular motion onto a diameter results in Simple Harmonic Motion (SHM).
The acceleration in SHM is given by $a = -\omega^2 x$. The magnitude of the acceleration is maximum at the extreme positions.
When the projection P touches point M, it is at an extreme position of its motion.
At this point, the displacement is maximum, equal to the amplitude $A$, which is the radius of the circle, $A = 0.36$ m.
The magnitude of the acceleration is maximum: $a_{max} = \omega^2 A$.
First, we calculate the angular velocity $\omega$. The point covers $\Delta\theta = 30^\circ$ in $\Delta t = 0.1$ s.
Converting the angle to radians: $\Delta\theta = 30^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{6}$ rad.
The angular velocity is $\omega = \frac{\Delta\theta}{\Delta t} = \frac{\pi/6}{0.1} = \frac{10\pi}{6} = \frac{5\pi}{3}$ rad/s.
Now, calculate the maximum acceleration:
$a_{max} = \omega^2 A = \left(\frac{5\pi}{3}\right)^2 \times 0.36 = \frac{25\pi^2}{9} \times 0.36$.
$a_{max} = 25\pi^2 \times 0.04 = \pi^2$.
Using $\pi \approx 3.14159$, we find $\pi^2 \approx 9.87$.
The restoration force per unit mass is $a_{max} = 9.87$ N/kg, or simply 9.87 N.