Question

A motorcycle, which has an initial linear speed of 9.8 m/s, decelerates to a speed of 2.2 m/s in 3.4 s. Each wheel has a radius of 0.65 m and is rotating in a counterclockwise (positive) directions. What is (a) the constant angular acceleration (in rad/s2) and (b) the angular displacement (in rad) of each

Answer 1

Answer:

The angular acceleration and angular displacement is $-2.24\ m/s^2$ and 31.34 rad .

Explanation:

Given :

Initial linear speed , u = 9.8 m/s .

Final speed , v = 2.2 m/s .

Time taken , t = 3.4 s .

Radius of wheel , r = 0.65 m .

So , decelerates of wheel is given by :

$v-u=at\\2.2-9.8=a\times 3.4\\a=-2.24\ m/s^2$

Therefore , angular velocity is given by :

$\omega=\dfrac{a}{r}\\\\\omega=\dfrac{-2.24}{0.65}\\\\\omega =-3.45\ rad/s^2$

Now , linear displacement is :

$s=ut+\dfrac{at^2}{2}\\\\s=9.8\times 3.4+\dfrac{-2.24\times 3.4^2}{2}\\\\s=20.37\ m$

Therefore , angular displacement is :

$d=\dfrac{s}{r}\\\\d=\dfrac{20.37}{0.65}\\\\d=31.34\ rad$

Hence , this is the required solution .