Answer:
α = 4.44*10^-5 rev/s^2
ar = 1.088*10^{-3}m/s^2
at = 1.88*10^{-4}m/s^2
Explanation:
T of ind the angular acceleration you can use the following formula from an accelerated harmonic motion:
[tex]\omega=\omega_o+\alpha t\\\alpha=\frac{\omega-\omega_o}{t}[/tex]
w: angular velocity
w_o: initial angular velocity
t: time
α: angular acceleration
In this case you have that the cylinder increase its angular speed 1.0rev/min until 12min. The angular acceleration must be computed for 6 min. Hence, by replacing you obtain:
[tex]\omega=1.0\frac{rev}{min}*\frac{1min}{60s}=0.016rev/s\\\alpha=\frac{0.016rev/s}{6(60s)}=4.44*10^{-5}\frac{rev}{s^2}[/tex]
hence, the angular acceleration is 4.44*10^-5 rev/s^2
the linear and tangential acceleration are calculated by using the formulas:
[tex]a_t=\alpha r\\\\a_r=\frac{v^2}{r}[/tex]
where v is the speed of the border of the cylinder, that is for r=8.5m/2=4.25m. By replacing you obtain:
[tex]a_t=(4.44*10^{-5}rev/s^2)(4.25m)=1.88*10^{-4}m/s^2\\\\v=\omega r=(0.016rev/s)(4.25m)=0.068m/s\\\\a_r=\frac{(0.068m/s)^2}{(4.25m)}=1.088*10^{-3}m/s^2[/tex]
hence, the radial acceleration is 1.088*10^{-3}m/s^2 and the tangential acceleration is 1.88*10^{-4}m/s^2
