The problem asks for the total number of revolutions a flywheel completes while its angular speed increases over a specific time period.
First, convert the initial and final angular speeds from revolutions per minute (rpm) to radians per second (rad/s), as physics calculations require SI units.
Determine the angular acceleration ($\alpha$) using the formula $\alpha = \frac{\omega_f - \omega_0}{t}$.
Use the kinematic equation for angular displacement: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Convert the total angular displacement ($\theta$) from radians to revolutions. Since $1 \text{ revolution} = 2\pi \text{ radians}$.
Therefore, the flywheel completes 150 revolutions.
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
Bob B of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10\text{ m}$, as shown in the figure. Point mass A of mass $m$ travelling horizontally with speed $10\text{ ms}^{-1}$ hits bob B elastically. The bob B rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\text{ ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is :
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is $L_A$ and $L_B$ computed about points A and B, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is :
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
