Consider a 20 kg uniform circular disc of radius 0.2 m. It is pin supported at its centre and is at

Q: Why do we use $I = \frac{1}{2}MR^2$ for the moment of inertia?

Because the object is a uniform solid circular disc rotating about its central perpendicular axis. This is the standard formula derived by integrating the mass distribution of a solid disc.

Q: How is the torque calculated here?

Torque is the cross product of the position vector and the force vector. Since the string is wrapped around the periphery, the force is applied tangentially. The angle between the radius and the force is 90 degrees, making the torque simply $\tau = F \times R$.

Q: Why is the initial angular velocity $\omega_0$ zero?

The problem explicitly states that the disc is 'at rest initially', which means it has no initial rotational speed.

Q: What is the relationship between angular displacement and the number of revolutions?

One complete revolution corresponds to an angular displacement of $2\pi$ radians (or $360^\circ$). Therefore, to find the total number of revolutions, you divide the total angular displacement in radians by $2\pi$.

Q: Why did we round 19.90 to 20?

The question explicitly asks for the value of $n$ 'to the nearest integer'. Since 19.90 is closer to 20 than to 19, we round up to 20.

Question

Consider a $20	ext{ kg}$ uniform circular disc of radius $0.2	ext{ m}$. It is pin supported at its centre and is at rest initially. The disc is acted upon by a constant force $F = 20	ext{ N}$ through a massless string wrapped around its periphery as shown in the figure.
Suppose the disc makes $n$ number of revolutions to attain an angular speed of $50	ext{ rad s}^{-1}$. The value of $n$ to the nearest integer, is .......... .
(Given, in one complete revolution, the disc rotates by $6.28	ext{ rad}$.)

Sigma Percentile · Accepted Answer

### Visualizing the Rotational Dynamics

Imagine a heavy, uniform circular disc pinned perfectly at its center so it can spin freely. A string is wrapped tightly around its outer edge, and a constant force of $20	ext{ N}$ is pulling on it. This force is going to make the disc spin faster and faster. Our goal is to find out exactly how many revolutions the disc makes by the time it reaches a blistering angular speed of $50	ext{ rad/s}$.

### The Turning Effect: Torque

First, we need to understand how much "turning force" is being applied. In physics, this is called torque ($	au$). Since the force is applied tangentially at the edge of the disc, the torque is simply the product of the force and the radius of the disc.

$$	au = F 	imes R$$

Substituting our given values:

$$	au = 20	ext{ N} 	imes 0.2	ext{ m} = 4	ext{ N-m}$$

### Resistance to Rotation: Moment of Inertia

Just as mass resists linear acceleration, the moment of inertia ($I$) resists angular acceleration. For a uniform solid disc rotating about its central axis, the formula is:

$$I = \frac{1}{2}MR^2$$

Plugging in the mass ($20	ext{ kg}$) and the radius ($0.2	ext{ m}$):

$$I = \frac{1}{2}(20)(0.2)^2 = 10 	imes 0.04 = 0.4	ext{ kg-m}^2$$

### Finding the Angular Acceleration

Now we can use Newton's Second Law for Rotation, which states that torque equals the moment of inertia times angular acceleration ($\alpha$).

$$	au = I\alpha \implies \alpha = \frac{	au}{I}$$

$$\alpha = \frac{4}{0.4} = 10	ext{ rad/s}^2$$

### Rotational Kinematics

We know the disc starts from rest ($\omega_0 = 0$) and reaches a final angular velocity ($\omega_f = 50	ext{ rad/s}$). We want to find the total angle it swept through ($\Delta	heta$). We can use the third equation of rotational kinematics:

$$\omega_f^2 = \omega_0^2 + 2\alpha\Delta	heta$$

Substituting our known values:

$$(50)^2 = 0^2 + 2(10)\Delta	heta$$

$$2500 = 20\Delta	heta$$

$$\Delta	heta = \frac{2500}{20} = 125	ext{ rad}$$

### Calculating the Number of Revolutions

We have the total angular displacement in radians, but the question asks for the number of revolutions ($n$). Since one complete revolution is $2\pi$ radians (given as $6.28	ext{ rad}$), we divide the total angle by the angle per revolution:

$$n = \frac{\Delta	heta}{2\pi} = \frac{125}{6.28} \approx 19.90$$

**The Final Answer:**
The question specifically asks for the nearest integer. Rounding $19.90$ gives us our final answer: **20 revolutions**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$τ = F \times R$

$I = \frac{1}{2} M R^{2}$

$α = \frac{τ}{I}$

$ω_{f}^{2} = ω_{0}^{2} + 2 α Δ θ$

$Δ θ = 125 rad$

$n = \frac{Δ θ}{2 π}$

$n \approx 20$

The Sigma Insight: Torque and Angular Momentum

Visualizing the Rotational Dynamics

The Turning Effect

Resistance to Rotation

Finding the Angular Acceleration

Rotational Kinematics

Calculating the Number of Revolutions

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Enter Numerical Value:

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; } τ=F×R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=21​MR2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=Iτ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ωf2​=ω02​+2αΔθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δθ=125 rad

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2πΔθ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n≈20