The angular speed of a motor wheel is increased from `600"rpm"` to `12 – InterviewSolution

InterviewSolution

1.	The angular speed of a motor wheel is increased from `600"rpm"` to `1200"rpm"` in `4` seconds. Calculate how many revolutions does the engine make during this time.
Answer» Using `omega=omega_(0)+alpha t` `alpha=(omega-omega_(0))/(t)` `omega_(0)=600"rpm"` `=(2pixx600)/(60)=20pi rad s^(-1)` `omega=1200"rpm"` `=(2pixx1200)/(60)=40pi rad s^(-1)` `alpha=(20pi)/(4)=5pi rad s^(-2)` again using, `theta=omega_(0)t+(1)/(2)alpha t^(2)` `theta=20pixx4+(1)/(2)xx5pixx16` `theta=80pi+40pi=120pi` Number of revolutions `=(theta)/(2pi)=(120pi)/(2pi)=60` revolutions

Discussion

No Comment Found

Related InterviewSolutions

Calculate the velocity with which a solid cylinder rolls down an inclined plane of height `10m` with slipping (take `g=10m//s^(2)`)
A sphere and a cylinder have the same radius and the same mass. They start from rest at the top of an incline. Which reaches the bottom first? Which has the greater speed at the bottom? Which has the greater total kinetic energy of the bottom ? Which has the greater rotational `KE`?
A uniform ring rolls down an inclined plane without slipping. If it reaches the bottom of a speed of `2m//s`, then calculate the height of the inclined plane (use `g=10m//s^(2)`)
A torque of `5Nm` is applied on a particle for `2` seconds, calculate the change in its angular momentum.
What is the angular acceleration of a particle moving with constant angular velocity ?
What is the angular acceleration of a particle if the angular velocity of a particle becomes `4` times of its initial angular velocity `1` rad s in `2` seconds
What is the angular momentum of a particle at rest ? Hint. `vecv=0`.
A wheel is rotating with an angular velocity of `3 rad s^(-1)`. If the angular acceleration is `2 rad s^(-1)` then calculate its angular velocity after `5` second.
A particle of mass `1kg` is moving about a circle of radius `1m` with a speed of `1m//s`. Calculate the angular momentum of the particle.
A force `(hati-2hatj+3hatk)` acts on a particle of position vector `(3hati+2hatj+hatk)`. Calculate the `hatj^(th)` component of the torque acting on the particle.