A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in

Q: Why is the moment of inertia about the diameter $\frac{MR^2}{4}$ and not $\frac{MR^2}{2}$?

The formula $\frac{MR^2}{2}$ is for an axis perpendicular to the plane of the disc passing through its center. Using the perpendicular axis theorem ($I_z = I_x + I_y$), since $I_x = I_y$ for a uniform disc, $2I_{\text{diameter}} = \frac{MR^2}{2}$, which gives $I_{\text{diameter}} = \frac{MR^2}{4}$.

Q: Why do we use the parallel axis theorem here?

The axis of rotation (stick AB) is parallel to the diameter of the coin and is shifted by a distance equal to the radius $R$. The parallel axis theorem ($I = I_{cm} + Md^2$) is the perfect tool for this.

Q: How do we convert rotations per second to rad/s?

One complete rotation equals $2\pi$ radians. Therefore, to convert rotations per second (rps) to radians per second (rad/s), you simply multiply by $2\pi$.

Q: Why is the initial angular velocity zero?

The problem states that 'The system is initially at rest', which means it is not rotating at the beginning. Hence, $\omega_i = 0$.

Q: What if the stick was perpendicular to the plane of the coin?

If the stick was perpendicular to the plane at the edge, the central parallel axis would be the one with $I = \frac{MR^2}{2}$. Using the parallel axis theorem, the new moment of inertia would be $\frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2$.

Question

A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5 s, is close to

Options:

(A) $4.0 	imes 10^{-6}$ N-m

(B) $2.0 	imes 10^{-5}$ N-m

(C) $1.6 	imes 10^{-5}$ N-m

(D) $7.9 	imes 10^{-6}$ N-m

Sigma Percentile · Accepted Answer

### The Spinning Coin: Mastering Torque and Moment of Inertia

Imagine a metal coin, which is basically a flat uniform disc, glued to a thin stick. The stick is touching the edge of the coin, acting as a tangent in the exact same plane as the coin itself. The entire setup is going to spin around this stick.

To figure out the constant torque needed to spin it, we must use the rotational equivalent of Newton's second law:

$$	au = I\alpha$$

So, our mission is to find these two critical pieces of the puzzle: the moment of inertia $I$ about the stick, and the angular acceleration $\alpha$.

### The Moment of Inertia Puzzle

Let's tackle the moment of inertia first. We know the inertia about an axis passing straight through the center, perpendicular to the coin, is $\frac{MR^2}{2}$. Using the perpendicular axis theorem ($I_z = I_x + I_y$), the inertia about any diameter is exactly half of that, which gives us:

$$I_{	ext{diameter}} = \frac{MR^2}{4}$$

But our axis of rotation is the stick, which is parallel to the diameter and shifted by a distance equal to the radius $R$. By applying the parallel axis theorem, we add $MR^2$ to the diameter's inertia.

$$I_{AB} = I_{	ext{diameter}} + MR^2 = \frac{MR^2}{4} + MR^2 = \frac{5}{4}MR^2$$

### Kinematics of Rotation

Now for the angular acceleration. The coin starts from rest, so the initial angular velocity $\omega_i$ is zero. It reaches a final speed of $25$ rotations per second. Since one rotation is $2\pi$ radians, the final angular velocity is:

$$\omega_f = 25 	imes 2\pi = 50\pi 	ext{ rad/s}$$

Dividing this change by the time of $5$ seconds gives us an angular acceleration of:

$$\alpha = \frac{50\pi}{5} = 10\pi 	ext{ rad/s}^2$$

### Bringing It All Together

We have our tools ready! Let's substitute the moment of inertia and angular acceleration into our torque equation. We must be careful to convert the mass to kilograms and the radius to meters. So, mass is $5 	imes 10^{-3}$ kg, and radius is $10^{-2}$ m.

$$	au = \left(\frac{5}{4}MR^2ight) \alpha$$

$$	au = \frac{5}{4} 	imes (5 	imes 10^{-3}) 	imes (10^{-2})^2 	imes 10\pi$$

Let's crunch the numbers. Multiplying these values, we get approximately $1.96 	imes 10^{-5}$ N-m. Rounding this off, our final answer is:

**$	au \approx 2.0 	imes 10^{-5} 	ext{ N-m}$**

Think about this: what if the stick was attached to the edge but perpendicular to the coin's surface? The moment of inertia would change to $\frac{3}{2}MR^2$, completely altering the required torque. Always visualize the axis carefully!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$The Master Equation$

$Moment of Inertia about Diameter$

$Moment of Inertia about Tangent$

$Calculating Angular Acceleration$

$Substituting the Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Torque and Angular Momentum

The Spinning Coin

The Moment of Inertia Puzzle

Kinematics of Rotation

Bringing It All Together

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Master Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia about Diameter

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia about Tangent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Angular Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward