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Chapter 5: Problem 615

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

Expert verified
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

01

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 = \( \frac{1500}{300} \)
02

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.
03

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.
04

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)
05

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.

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Most popular questions from this chapter

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

The angular momentum of a wheel changes from \(2 \mathrm{~L}\) to $5 \mathrm{~L}$ in 3 seconds what is the magnitudes of torque acting on it? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The magnitude of torque acting on the disc is \(\\{\mathrm{A}\\} \mathrm{MgR}\) \(\\{\mathrm{B}\\} \mathrm{MgR} \sin \theta\) \(\\{\mathrm{C}\\}[(2 \mathrm{MgR} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(\mathrm{MgR} \sin \theta) / 3]\)

A solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

A uniform rod of length \(\mathrm{L}\) is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? \(\\{\mathrm{A}\\} \sqrt{(2 \mathrm{~g} \mathrm{~L})}\) \(\\{\mathrm{B}\\} 2 \sqrt{(\mathrm{gL})}\) \(\\{C\\} \sqrt{(6 g L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 g \mathrm{~L})}\)
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