Question

The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?

Answer 1

Answer:

Torque = 8.38Nm

Explanation:

Time= 8.00s

angular speed (w) =400 rpm

Moment of inertia (I)= 1.60kg.m2 about its rotation axis

We need to convert the angular speed from rpm to rad/ sec for consistency

2PI/60*n = 0.1047*409 = 41.8876 rad/sec

What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

Then we need to use the formula below for our torque calculation

from basic equation T = J*dω/dt ...we get

Where : t= time in seconds

W= angular velocity

T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

Therefore, constant torque that is required is 8.38 Nm

Answer 2

Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.

Given-

Inertia of the flywheel is 1.60 kg m squared.

Angular speed of the flywheel $n$ is 400 rpm. Convert it into the rad/sec, we get,

$\omega =\dfrac{2\pi }{60} \times n$

$\omega =\dfrac{2\pi }{60} \times 400$

$\omega = 41.89$

Thus, the angular speed of the flywheel $\omega$ is 41.89 rad/sec.

When a torque $\tau$ is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia $I$. Mathematically,

$\tau=\dfrac{\Delta \omega }{\Delta t} \times I$

$\tau=\dfrac{ 41.89 }{8} \times 1.6$

$\tau=8.38$

Hence, the required constant torque is 8.38 N-m.

to know more about the torque, follow the link below-

https://brainly.com/question/6855614