Question

A motorcycle, which has an initial linear speed of 6.3 m/s, decelerates to a speed of 1.9 m/s in 4.6 s. Each wheel has a radius of 0.68 m and is rotating in a counterclockwise (positive) direction.a) the constant angular acceleration (in rad/s2) andb) the angular displacement (in rad) of each wheel?

Answer 1

Answer:

Explanation:

Given

initial velocity of motorcycle $u=6.3\ m/s$

final velocity $v=1.9\ m/s$

time taken $t=4.6\ s$

radius of wheel $r=0.68\ m$

using v=u+at

where a=acceleration

$1.9=6.3+a\times 4.6$

$a=-0.956\ m/s^2$

angular acceleration $\alpha =\frac{a}{r}$

$\alpha =\frac{-0.956}{0.68}=-1.406\ rad/s^2$

$Displacement=average\ velocity\times time$

$Displacement=\frac{u+v}{2}\times t$

$Displacement=\frac{1.9+6.3}{2}\times 4.6=18.86\ m$

Angular displacement $\theta =\frac{Linear\ displacement}{radius}$

Angular displacement $\theta =\frac{18.86}{0.68}=27.73\ rad$

Answer 2

Answer:

Explanation:

initial velocity, u = 6.3 m/s

final speed, v = 1.9 m/s

time, t = 4.6 s

radius, r = 0.68 m

(a)

initial angular velocity, ωo = u / r = 6.3 / 0.68 = 9.26 rad/s

final angular velocity, ω = v / r = 1.9 / 0.68 = 2.79 rad/s

Let α be the angular acceleration

Use first equation of motion

ω = ωo + αt

2.79 = 9.26 + α x 4.6

α = - 1.41 rad/s²

(b) Let Ф be the angular displacement

Use third equation of motion

ω² = ωo² + 2 αФ

2.79² = 9.26² - 2 x 1.41 x Ф

Ф = 27.6 rad