Explanation:
It is given that,
Length of the bicycle crank, l = 15 cm = 0.15 m
Diameter of sprocket, d = 21 cm
Radius, r = 0.105 m
Initial angular velocity, [tex]\omega_i=62\ rpm=6.49\ rad/s[/tex]
Final angular velocity, [tex]\omega_f=95\ rpm=9.94\ rad/s[/tex]
Time, t = 12 s
(a) The relation between the tangential acceleration and the angular acceleration. It is given by :
[tex]a=\alpha r[/tex]
[tex]a=\dfrac{\omega_f-\omega_i}{t}\times r[/tex]
[tex]a=\dfrac{9.94-6.49}{12}\times 0.105[/tex]
[tex]a=0.0301\ m/s^2[/tex]
(b) Using third equation of kinematics as :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha}[/tex]
[tex]\theta=\dfrac{9.94^2-6.49^2}{2\times \dfrac{9.94-6.49}{12}}[/tex]
[tex]\theta=98.58\ rad[/tex]
Let l is the length of the chain,
[tex]l=r\times \theta[/tex]
[tex]l=0.105\times 98.58[/tex]
l = 10.35 m
Hence, this is the required solution.
