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
[tex]\theta = \omega_0 t + \alpha t^2/2[/tex]
[tex]17.3 = 2.1*5.41 + \alpha * 5.41^2/2[/tex]
[tex]17.3 = 11.361 + 14.63 \alpha[/tex]
[tex] \alpha = \frac{17.3 - 11.361}{14.63} = 0.406 rad/s^2[/tex]
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.
