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 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)

Short 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-by-step solution

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 = \( \frac{1500}{300} \)

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.

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.

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)

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.