Physics, 12.11.2019 05:31 MalikaJones

Aheavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. the flywheel is assumed to be at rest at time t=0 in parts a and b of this problem.

part a

find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.

express your answer in terms of ω1 and α.

part b

find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.

express your answer in terms of some or all of the following: ω1, α, and t1.

part c

assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. at this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).

express your answer in terms of some or all of the following: ω1, α, and t1.

Answers: 2

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Answer from: tommyaberman

(A) The time taken by the flywheel for the acceleration is, $\dfrac{\omega_{1}}{\alpha}$ .

(B) The angle turned by the flywheel is $\dfrac{1}{2} \alpha t^2_{1}$ .

Given data:

The magnitude of angular acceleration is, $\alpha$ .

Final angular speed is, $\omega_{1}$ .

Time taken is, $t_{1}$ .

(A)

Apply the first rotational equation of motion as,

$\omega_{1}=\omega' + \alpha t_{1}$

Since, the flywheel was initially at rest. So, $\omega'=0$ . Then,

$\omega_{1}=0 + \alpha \times t_{1}\\t_{1}=\dfrac{\omega_{1}}{\alpha}$

Thus, the time taken for the acceleration is, $\dfrac{\omega_{1}}{\alpha}$ .

(B)

Now apply the second rotational equation of motion to obtain the angle turned by the flywheel as,

$\theta_{1}=\omega ' \times t_{1}+\dfrac{1}{2} \alpha t^2_{1}$

Substituting the values as,

$\theta_{1}=0 \times t_{1}+\dfrac{1}{2} \alpha t^2_{1}\\\theta_{1}=\dfrac{1}{2} \alpha t^2_{1}$

Thus, the angle turned by the flywheel is $\dfrac{1}{2} \alpha t^2_{1}$ .

$\omega_{1}=\omega' + \alpha t_{1}\\\omega_{1}=0 + \alpha t_{1}\\t_{1}=\dfrac{\omega_{1}}{\alpha}$

Since, brake is applied. Which means the flywheel is stopped finally. So, applying the third rotational equation of motion as,

$\omega_{2}=\omega_{1}+(-5 \alpha)t'$

Here, $\omega_{2}$ is the final speed after brake is applied. And its value is $\omega_{2} =0$ .

Solving as,

$0=\omega_{1}-5 \alpha t\\t'=\dfrac{\omega_{1}}{5 \alpha } \\t'=\dfrac{1}{ 5} \times \dfrac{\omega_{1}}{ \alpha } \\t'=\dfrac{t_{1}}{ 5}$

Thus, the time taken by the flywheel to stop after the brake is applied is $\dfrac{t_{1}}{ 5}$ .

Learn more about the rotational kinematics here:

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Answer from: schoolemailonlyno1

Part a)

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

$t = \frac{t_1}{5}$

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

$\omega_f = \omega_i + \alpha t$

$\omega_1 = 0 + \alpha t_1$

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

angular displacement is given as

$\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2$

$\theta = 0 + \frac{1}{2}\alpha t_1^2$

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

As we know that the angular deceleration produced by the brakes is given as

$\alpha_d = - 5\alpha$

now we have

$\omega_f = \omega _i + \alpha t$

$0 = \omega_1 - 5 \alpha_1 t$

$t = \frac{\omega_1}{5 \alpha}$

As we know that

$t_1 = \frac{\omega_1}{\alpha}$

so we have

$t = \frac{t_1}{5}$

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