Question

A rotating wheel with diameter 0.800 m is speeding up with constant angular acceleration. The speed of a point on the rim of the wheel increases from 3.00 m/s to 6.00 m/s while the wheel turns through 4.00 revolutions.
Required:
What is the angular acceleration of the wheel?

Answer 1

Answer:

The angular acceleration of the wheel is 3.357rad/s²

Explanation:

Given;

Diameter of the wheel, d = 0.8 m

radius of the wheel, r = 0.4 m

initial velocity of the wheel, u = 3 m/s

final velocity of the wheel, v = 6 m/s

number of revolutions of the wheel, N = 4 rev.

Convert the linear velocity in m/s  to angular velocity in rad/s;

v = ωr

ω = v / r

initial angular velocity, $\omega_i = \frac{3}{0.4} = 7.5 \ rad/s$

final angular velocity, $\omega_f= \frac{6}{0.4} = 15\ rad/s$

The angular acceleration of the wheel is calculated as;

$\omega _f^2 = \omega_i^2 + 2 \alpha \theta$

where;

α is the angular acceleration (rad / s²)

θ is the angular rotation (rad)

θ = Number of revolutions x 2π rad/rev

  =  4 rev. x 2π rad/rev

  = 25.136 rad.

$\omega _f^2 = \omega_i^2 + 2 \alpha \theta\\\\2 \alpha \theta = \omega _f^2 - \omega_i^2\\\\\alpha = \frac{\omega _f^2 - \omega_i^2}{2 \theta} \\\\\alpha = \frac{15^2 -7.5^2}{2*25.136} \\\\\alpha = 3.357 \ rad/s^2$

α = 3.357 rad/s²

Therefore, the angular acceleration of the wheel is 3.357rad/s²