Question

A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude 5.0 rad/s2. How many revolutions does it make while slowing down before coming to rest

Answer 1

Answer:

It takes 6.37 revolutions to stop.

Explanation:

The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0$\frac{rad}{s^{2}}$.Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:

$\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta$

with $\omega$ the final angular velocity (is zero because the wheel comes to rest), $\omega_{i}$ the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :

$\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha}$

$\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad$

The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:

$40rad=\frac{40rad}{2\pi\frac{rad}{rev}}$

$40rad=6.37 rev$

Answer 2

Number of revolution take place is 6.4 revolution

Given that;

Velocity of rotating wheel = 20 rad/s

Acceleration of magnitude = 5.0 rad/s²

Find:

Number of revolution take place

Computation:

Using Third equation of motion;

v² - u² = 2as

20² - 0² = 2(5)(s)

400 = (10)(s)

Total distance = 40 rad

Number of revolution take place = Total distance / 2π

Number of revolution take place = 40 / 2(3.14)

Number of revolution take place = 40 / 6.28

Number of revolution take place = 6.369

Number of revolution take place = 6.4 revolution

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