The angular velocity of the rear wheel of a stationary exercise bike is 5.40 rad/s at time t = 0.000 sec, and its angular acceleration is constant and equal to 1.50 rad/s^2. A particular spoke coincides with the +x axis at time t = 0.000 sec. (a) What angle (in rev) does this spoke make with the +x axis at time t = 4.00 s? (b) What is the angular velocity (rev/s) at this time?

Respuesta :

Answer:

The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.

Explanation:

Given that,

Angular velocity = 5.40 rad/s

Time t = 0.000 sec

Angular acceleration = 1.50 rad/s^2

(a). We need to calculate the angle at time t = 4.00 s

Using formula for angle

[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t^2[/tex]

Where, [tex]\omega_{0}[/tex]=angular velocity

[tex]\alpha[/tex]=angular acceleration

t = time

Put the value into the formula

[tex]\theta=5.40\times4.00+\dfrac{1}{2}\times1.50\times(4.00)^2[/tex]

[tex]\theta=33.6\ rad[/tex]

[tex]\theta=\dfrac{33.6}{2\pi}\ rad[/tex]

[tex]\theta=5.35\ rev[/tex]

(b). We nee to calculate the angular velocity at 4.00 s

Using formula of angular velocity

[tex]\omega=\omega_{0}+\alpha t[/tex]

[tex]\omega =5.40+1.50\times4.00[/tex]

[tex]\omega=11.4\ rad/s[/tex]

[tex]\omega=\dfrac{11.4}{2\pi}\ rad/s[/tex]

[tex]\omega=1.81\ rev/s[/tex]

Hence, The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.