Explanation:
Given that,
A motorcycle, which has an initial linear speed of 5.0 m/s, decelerates to a speed of 3.5 m/s in 4.5 s. Each wheel has a radius of 0.60 m and is rotating in a counterclockwise (positive) directions.
Angular acceleration, [tex]\alpha =\dfrac{\omega_2-\omega_1}{t}[/tex]
[tex]\alpha =\dfrac{\dfrac{v_2}{r}-\dfrac{v_1}{r}}{t}[/tex]
Put all the values,
[tex]\alpha =\dfrac{\dfrac{3.5}{0.6}-\dfrac{5}{0.6}}{4.5}\\\\=-0.56\ rad/s^2[/tex]
Angular displacement,
[tex]\theta=\dfrac{\omega_2^2-\omega_1^2}{2\alpha}\\\\\theta=\dfrac{(\dfrac{v_2}{r})^2-(\dfrac{v_1}{r})^2}{2\alpha}\\\\\theta=\dfrac{(\dfrac{3.5}{0.6})^2-(\dfrac{5}{0.6})^2}{2\times (-0.56)}\\\\=31.62\ rad[/tex]
Hence
Answer:
(a) The angular acceleration is - 0.56 rad/s^2.
(b) The angular displacement is 31.6 rad.
Explanation:
initial velocity, u = 5 m/s
final velocity, v = 3.5 m/s
radius, r = 0.6 m
time, t = 4.5 s
initial angular velocity, wo = u/r = 5/0.6 = 8.33 rad/s
final angular velocity, w = v/r = 3.5 / 0.6 = 5.83 rad/s
(a) Use the first equation of motion to fine the angular acceleration.
[tex]w = w_o + \alpha t \\\\5.83 = 8.33 + \alpha \times 4.5\\\\\alpha = - 0.56 rad/s^2[/tex]
(b) Use third equation of motion to find the angular displacement
[tex]w^2 = w_0^2 + 2\alpha \theta \\\\5.83^2 =8.833^2 - 2 \times 0.56\times \theta \\\\\theta =31.6 rad[/tex]
