Question
A heavy flywheel is accelerated (rotationally) by a motor thatprovides constant torque and therefore a constant angularacceleration Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta. The flywheel is assumed to be at rest at timet = 0 in Parts A and B of this problem.

a) Find the time t_1 it takes to accelerate the flywheel to omega_1 if the angular acceleration is Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.
Express your answer in terms ofomega_1 and Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.
b)
Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.
Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.
Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.
Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.
c)
Assume that the motor has accelerated thewheel up to an angular velocity omega_1 with angular acceleration Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta in time t_1. At this point, the motor is turned off and a brake isapplied that decelerates the wheel with a constant angularacceleration of -5. Find t_2, the time it will take the wheel to stop after the brakeis applied (that is, the time for the wheel to reach zero angularvelocity).
Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.
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Answer #1
Concepts and reason

The main concept used to solve the problem is rotational kinematics equations.

Initially, use the constant angular equation from rotational kinematics to calculate the time taken by the flywheel to attain an angular acceleration. Later, use constant angular equation for angular displacement to calculate the angle through which the flywheel turned during the time interval. Finally, use the constant angular equation from rotational kinematics to find the time taken by the wheel to stop.

Fundamentals

The constant angular acceleration equation is expressed as follows:

ω1=ω0+αt{\omega _1} = {\omega _0} + \alpha t

Here, ω1{\omega _1} is the angular velocity of the flywheel at time t1{t_1}, ω0{\omega _0} is the angular velocity of the flywheel at time 0, t is the time, and α\alpha is the angular acceleration.

The equation which relates the constant angular acceleration with the angular displacement is expressed as follows:

θ=ω0t+12αt2\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}

Here, θ\theta is the angular displacement, t is the time, α\alpha is the angular acceleration, and ω0{\omega _0} is the angular velocity of the flywheel at time 0.

(a)

Calculate the time taken by the flywheel to attain an angular acceleration.

The constant angular acceleration equation is expressed as follows:

ω1=ω0+αt{\omega _1} = {\omega _0} + \alpha t

Here, ω1{\omega _1} is the angular velocity of the flywheel at time t1{t_1}, ω0{\omega _0} is the angular velocity of the flywheel at time 0, t is the time, and α\alpha is the angular acceleration.

Substitute 0 rad/s for ω0{\omega _0} in expression ω1=ω0+αt{\omega _1} = {\omega _0} + \alpha t and solve for t1{t_1}.

ω1=(0)+αt1t1=ω1α\begin{array}{c}\\{\omega _1} = \left( 0 \right) + \alpha {t_1}\\\\{t_1} = \frac{{{\omega _1}}}{\alpha }\\\end{array}

(b)

Calculate the angular displacement.

The equation which relates the constant angular acceleration with the angular displacement is expressed as follows:

θ=ω0t+12αt2\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}

Here, θ\theta is the angular displacement, t is the time, α\alpha is the angular acceleration, and ω0{\omega _0} is the angular velocity of the flywheel at time 0.

Substitute 0 for ω0{\omega _0} in expression θ1=ω0t1+12αt12{\theta _1} = {\omega _0}{t_1} + \frac{1}{2}\alpha {t_1}^2.

θ1=(0rad/s)t1+12αt12θ1=12αt12\begin{array}{c}\\{\theta _1} = \left( {0\,{\rm{rad/s}}} \right){t_1} + \frac{1}{2}\alpha {t_1}^2\\\\{\theta _1} = \frac{1}{2}\alpha {t_1}^2\\\end{array}

(c)

Calculate the time taken by the flywheel to stop after the brakes applied.

The constant angular acceleration equation is expressed as follows:

ω1=ω0+αt{\omega _1} = {\omega _0} + \alpha t

Here, ω1{\omega _1} is the angular velocity of the flywheel at time t1{t_1}, ω0{\omega _0} is the angular velocity of the flywheel at time 0, t is the time, and α\alpha is the angular acceleration.

At the instant of the time just before the motor tuned off, the angular velocity ω0{\omega _0} is equal to zero.

Now, the initial conditions have change, in this case wheel is initially spinning and angular acceleration is different.

Substitute 0 for ω0{\omega _0}and 5α - 5\alpha for α\alpha in equation ω0=ω1+αt{\omega _0} = {\omega _1} + \alpha t.

(0)=ω1+(5α)t2t2=ω15α\begin{array}{c}\\\left( 0 \right) = {\omega _1} + \left( { - 5\alpha } \right){t_2}\\\\{t_2} = \frac{{{\omega _1}}}{{5\alpha }}\\\end{array}

Ans: Part a

The time taken by the flywheel to attain an angular acceleration is ω1α\frac{{{\omega _1}}}{\alpha }.

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