Question

A flywheel has a moment of inertia 0.5 kg-m
2
. What torque is required to
increase its velocity from 2 rad s
−1
to 18rad s
−1
in 8 s ?

(a) 1.0 N-m (b) 2.0 N-m (c) 1.5 N-m (d) 3.0 N-m​

Answer 1

Answer:0.65 π Nm

Explanation:ωf = ωi + αt

where:

ωi = 2π rad/s (initial angular velocity)

ωf = 15π rad/s (final angular velocity)

t = 10 s (time interval)

Rearranging the formula to solve for α, we get:

α = (ωf - ωi) / t

α = (15π rad/s - 2π rad/s) / 10 s

α = 1.3 π rad/s^2

Now, we can use the formula for torque to determine the amount of torque required to produce this angular acceleration:

τ = Iα

where:

I = 0.5 kg m^2 (moment of inertia)

τ = 0.5 kg m^2 * 1.3π rad/s^2

τ = 0.65π Nm

Therefore, the amount of torque required to increase the speed of the flywheel from 2π rad/s to 15π rad/s in 10 seconds is 0.65π Nm.

Answer 2

Answer:

1 N-m is the torque required to increase the velocity of the flywheel from 2 rad/s to 18 rad/s.

Explanation:

It is given that,

Moment of inertia, $I = 0.5kg-m^{2}$

Increase in velocity from $2rads^{-1}$ to $18rads^{-1}$

Time interval, Δt$=8 s$

We know that,

$Torque=I\alpha$

Torque is a measure of the force that can cause an object to rotate about an axis, that can be either center of mass or any fixed point.

$\alpha =$ Δω / t

τ = (Δω* I) / Δt

Here,

Δω  is the change in angular velocity i.e.  final angular velocity - initial angular velocity

τ is the torque applied

Δt is the time interval

I is the moment of inertia

The moment of inertia of a rotating rigid body is the sum of the product of mass of each point and square of its distance from the axis of rotations.

Δω $=18-2=16$

τ $=\frac{ 16*0.5}{8}=1$

Therefore, the torque required is $1 N-m.$

To know more about torque go to the following link.

https://brainly.in/question/2199910

To know more about moment of inertia go to the following link.

https://brainly.in/question/14617883

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