Problem 615 - JEE Physics

Question

A constant torque of \(1500 \mathrm{Nm}\) turns a wheel of moment of inertia \(300 \mathrm{~kg} \mathrm{~m}^{2}\) about an axis passing through its centre the angular velocity of the wheel after 3 sec will be.......... \(\mathrm{rad} / \mathrm{sec}\) \(\\{\mathrm{A}\\} 5\) \\{B \\} 10 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)

Step-by-Step Solution

Answer

The final angular velocity of the wheel after 3 seconds is 15 rad/s. Therefore, the correct answer is \( \boxed{\textbf{C } 15} \) rad/s.

Step 1: Calculate angular acceleration

To find out the angular acceleration (α), we can use the formula: α = torque/moment of inertia Given torque, T = 1500 Nm and moment of inertia, I = 300 kg m², the angular acceleration can be calculated as follows: α = \( \frac{1500}{300} \)

Step 2: Simplify the angular acceleration

Now we need to simplify the expression for angular acceleration. α = \( \frac{1500}{300} \) = 5 rad/s² The angular acceleration is 5 rad/s².

Step 3: Determine the initial angular velocity

We haven't been given any information about the initial angular velocity (ω₀) of the wheel in the question. So, let's assume the wheel is initially at rest which means the initial angular velocity ω₀ = 0 rad/s.

Step 4: Calculate the final angular velocity after 3 seconds

To find the final angular velocity (ω) after 3 seconds, we can use the formula: ω = ω₀ + αt We know ω₀ = 0 rad/s, α = 5 rad/s², and time t = 3 seconds. We can now plug in these values into the equation: ω = 0 + (5 × 3)

Step 5: Simplify the final angular velocity expression

Now, we will simplify the expression to calculate the final angular velocity: ω = 0 + (5 × 3) = 0 + 15 = 15 rad/s The final angular velocity of the wheel after 3 seconds is 15 rad/s. So, the correct answer is \( \boxed{\textbf{C } 15} \) rad/s.

Key Concepts

Moment of Inertia

Moment of inertia represents an object's resistance to changes in its rotational motion. Think of it as the rotational equivalent of mass in linear motion.
It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation.
For instance, a wheel with more mass located further from the axis will have a higher moment of inertia than one with mass concentrated near the axis.

• Symbol: Typically represented by the letter \( I \).
• Unit: The unit of moment of inertia is \( ext{kg} ext{ m}^2 \).
• Example: For a solid disc, the moment of inertia is \( I = \frac{1}{2} m r^2 \), where \( m \) is mass and \( r \) is radius.

In our problem, the wheel has a moment of inertia of \( 300 \, \text{kg m}^2 \), which influences how it responds to applied torques. This relationship is crucial for determining how the wheel accelerates.

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate.
It's like the rotational counterpart of linear force. Imagine pushing a door open; the further you push from the hinges, the easier it is to rotate.
Torque depends on the magnitude of the force, the distance from the pivot, and the angle at which the force is applied.

• Symbol: Typically denoted by \( \tau \).
• Unit: Measured in Newton meters (\( ext{Nm} \)).
• Equation: \( \tau = r \times F \times \sin(\theta) \), where \( r \) is the distance from the pivot point, \( F \) is the force, and \( \theta \) is the angle.

In the given problem, a constant torque of \( 1500 \, \text{Nm} \) is applied to the wheel, which helps it spin around its axis.

Angular Acceleration

Angular acceleration refers to how quickly an object's angular velocity changes. It tells us the rate at which an object speeds up or slows down its rotation.
Much like linear acceleration, angular acceleration is key to understanding rotational movement. In our context, it's influenced by both torque and the object's moment of inertia.

• Symbol: Denoted by \( \alpha \).
• Unit: Its unit is radians per second squared (\( ext{rad/s}^2 \)).
• Relationship: Given by \( \alpha = \frac{\tau}{I} \), which means it's directly proportional to the applied torque and inversely proportional to the moment of inertia.

For the problem, we find \( \alpha = \frac{1500}{300} = 5 \, \text{rad/s}^2 \). This shows the wheel gains or loses angular velocity steadily due to the applied torque.

Angular Velocity

Angular velocity tells us how fast something spins around its axis. It plays a similar role for rotating objects that linear velocity plays for objects moving along a straight path.
It describes how quickly the angle changes with time and helps us predict how fast the object rotates after a set period.

• Symbol: Usually represented by \( \omega \).
• Unit: The unit is radians per second (\( ext{rad/s} \)).
• Calculation: Calculated using the formula \( \omega = \omega_0 + \alpha t \), where \( \omega_0 \) is the initial angular velocity, \( \alpha \) is angular acceleration, and \( t \) is time.

In our problem, since the initial angular velocity \( \omega_0 \) is \( 0 \) and \( t = 3 \) seconds, the final angular velocity is \( 0 + (5 \times 3) = 15 \, \text{rad/s} \), reflecting the spin rate after the torque has acted over the given time.