Question

The wheel on an upside-down bicycle moves through 17.3 rad in 5.41 s. What is the wheel’s angular acceleration if its initial angular speed is 2.1 rad/s? Answer in units of rad/s 2 .

Answer 1

Answer:

0.406 rad/s2

Explanation:

We can use the following equation of motion to calculate the angular acceleration α in term of initial angular speed ω = 2.1 rad/s, time t = 5.41 and angles covered θ = 17.3 rad

$\theta = \omega_0 t + \alpha t^2/2$

$17.3 = 2.1*5.41 + \alpha * 5.41^2/2$

$17.3 = 11.361 + 14.63 \alpha$

$\alpha = \frac{17.3 - 11.361}{14.63} = 0.406 rad/s^2$

Answer 2

Answer:

0.406 rad/s^2.

Explanation:

Equating equations of linear motion to angular motion;

ω = θ/t

ωf^2 = ωi^2 + 2a*θ

θ = ωi*t + (1/2)*a*t^2

where,

ωi = initial angular velocity (rad/s),

ωf = final angular velocity (rad/s),

θ = angular distance (rad),

a = acceleration due to gravity (rad/s^2),

and t = time taken for motion.

θ = ωi*t + (1/2)*a*t^2,

17.3 = (2.1*5.41) +(1/2)*a*(5.41)^2

17.3 = 11.361 + 14.634*a

17.3 - 11.361 = 14.634*a

14.634*a = 5.939

a = 5.939/14.634

= 0.406 rad/s^2.