The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and cove

Q: Why is the restoring force per unit mass equal to acceleration?

By Newton's second law of motion, Force ($F$) equals mass ($m$) times acceleration ($a$). Therefore, rearranging the formula gives $F/m = a$. The force per unit mass is literally just the acceleration of the object.

Q: Why do we calculate acceleration specifically at point M?

The problem explicitly asks for the force when the projection $P$ touches $M$. Point $M$ lies on the circumference, making it the extreme position of the SHM path. At extreme positions, velocity is zero, but the restoring force (and thus acceleration) is at its maximum.

Q: How is the projection's acceleration related to the circular motion?

The maximum acceleration of the projection in SHM is exactly equal to the constant centripetal acceleration ($a_c = \omega^2 R$) of the reference particle moving in uniform circular motion. The projection's acceleration at any point is $a = -\omega^2 x$, and at the extreme ($x=R$), its magnitude is $\omega^2 R$.

Q: Why did we convert $30^\circ$ to radians?

In physics formulas involving angular velocity like $\omega = \theta/t$, the angle must strictly be in radians to yield standard SI units (rad/s). Using degrees would result in incorrect scaling when calculating linear acceleration.

Q: Is $\pi^2$ always exactly 9.87?

No, it is an approximation. Since $\pi \approx 3.14159$, squaring it gives $\pi^2 \approx 9.8696$, which rounds to $9.87$. It's a very common approximation used in physics, especially since it is numerically very close to the acceleration due to gravity $g$ ($9.8\text{ m/s}^2$).

Question

The point $A$ moves with a uniform speed along the circumference of a circle of radius $0.36	ext{ m}$ and covers $30^\circ$ in $0.1	ext{ s}$. 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

Options:

(A) $100	ext{ N}$

(B) $9.87	ext{ N}$

(C) $50	ext{ N}$

(D) $0.49	ext{ N}$

Sigma Percentile · Accepted Answer

### The Magic of Projections

Imagine you are standing above a rotating carousel, shining a flashlight straight down. If you place a small object on the edge of the carousel, its shadow will move back and forth along the floor in a straight line. This mesmerizing back-and-forth motion of the shadow is exactly what physicists call **Simple Harmonic Motion (SHM)**.

In our problem, the particle $A$ is moving smoothly along a circular path, and its perpendicular projection $P$ on the diameter $MN$ is executing SHM. This connection is not just a neat visual trick; it is one of the most powerful mathematical tools in physics, known as the **Phasor Approach**.

### Decoding the Kinematics

Before we jump into forces, we need to understand how fast our particle is spinning. The problem states that the particle covers an angular distance of $30^\circ$ in $0.1	ext{ s}$.

In physics, we must always work in radians. So, we convert $30^\circ$ to $\frac{\pi}{6}	ext{ rad}$. The angular velocity $\omega$ is simply the rate of change of angular displacement:

$$\omega = \frac{	heta}{t} = \frac{\pi/6}{0.1} = \frac{10\pi}{6} = \frac{5\pi}{3}	ext{ rad/s}$$

### The Core Concept: Force and Acceleration

The question asks for a slightly unusual quantity: the **restoring force per unit mass** when the projection $P$ touches point $M$.

By Newton's Second Law, $F = ma$, which means the force per unit mass is exactly equal to the acceleration:

$$\frac{F}{m} = a$$

Point $M$ is the extreme end of the diameter. In SHM, the extreme positions are where the particle momentarily stops and turns around. This is where the restoring force—and therefore the acceleration—is at its absolute maximum.

One of the beautiful symmetries of the phasor approach is that the maximum acceleration of the projection in SHM is exactly equal to the constant centripetal acceleration of the particle moving in the circle. Therefore:

$$a_{max} = \omega^2 R$$

### The Elegant Calculation

Now, we substitute our known values into the maximum acceleration formula. We have $\omega = \frac{5\pi}{3}	ext{ rad/s}$ and the radius $R = 0.36	ext{ m}$:

$$a_{max} = \left(\frac{5\pi}{3}ight)^2 	imes 0.36$$

Let's expand the squared term carefully:

$$a_{max} = \frac{25\pi^2}{9} 	imes 0.36$$

Notice how beautifully the numbers cancel out. Dividing $0.36$ by $9$ gives $0.04$. And multiplying $25$ by $0.04$ gives exactly $1$!

$$a_{max} = 25\pi^2 	imes 0.04 = 1 	imes \pi^2 = \pi^2$$

As a standard approximation in physics, $\pi^2 \approx 9.87$. Therefore, the restoring force per unit mass is **$9.87	ext{ N/kg}$**, which perfectly matches option (b).

Select Answer:

Visualized Solution

Visualizing the Motion

Restoring Force per Unit Mass

Calculating Angular Velocity $ω$

Setting up the Equation

Evaluating Maximum Acceleration

Conclusion

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

The Magic of Projections

Decoding the Kinematics

The Core Concept

The Elegant Calculation

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