Respuesta :
Answer:
The number of revolution that the wheel turn before it stops is [tex]\bf{62.7~rad}[/tex].
Explanation:
Given:
Initial angular velocity of the motor, [tex]\omega_{i} = 176~rev/min[/tex]
Final angular velocity of the motor, [tex]\omega_{f} = 0[/tex]
Value of constant angular deceleration, [tex]\alpha = 2.71~rad/s^{2}[/tex]
Consider that [tex]t[/tex] is the time taken by the motor to come to at rest.
The equation of motion of the motor is given by
[tex]\omega_{f} = \omega_{i} - \alpha t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
Substituting the values in equation (1),
[tex]t &=& \dfrac{\omega_{i}}{\alpha}\\ &=& \dfrac{176~(\dfrac{rev}{min})(\dfrac{2 \pi~rad}{1~rev})(\dfrac{1~min}{60~s})}{2.71~rad/s^{2}}\\ &=& 6.8~s[/tex]
Consider the angle of revolution is [tex]\theta[/tex].
The equation for revolution is given by
[tex]\theta = \omega_{i}t - \dfrac{1}{2}\alpha t^{2}~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Substituting the values in equation (2),
[tex]\theta = (176~(\dfrac{rev}{min})(\dfrac{2 \pi~rad}{1~rev})(\dfrac{1~min}{60~s}))(6.8~s) - \dfrac{1}{2}(2.71~rad/s^{2})(6.8~s)^{2}\\~~&=& 62.7~rad[/tex]
The revolutions the wheel turn before it finally comes to rest will be "62.7 rad".
Velocity and Acceleration
According to the question,
Initial angular velocity, [tex]\omega_i[/tex] = 176 rev/min
Final angular velocity, [tex]\omega_f[/tex] = 0
Constant angular deceleration, α = 2.71 rad/s²
By using the equation of motion,
→ [tex]\omega_f[/tex] = [tex]\omega_i[/tex] - αt
or,
t = [tex]\frac{\omega_i}{\alpha}[/tex]
By substituting the values,
= [tex]\frac{176\times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s} }{2.71}[/tex]
= 6.8 s
hence,
By using equation of revolution,
→ θ = [tex]\omega_i[/tex] t - [tex]\frac{1}{2}[/tex] αt²
By substituting the values,
= 176 × [tex]\frac{2 \ pi \ rad}{1 \ rev}[/tex] × [tex]\frac{1 \ min}{60 \ s}[/tex] - [tex]\frac{1}{2}[/tex] × 2.71 × (6.8)²
= 62.7 rad
Thus the response above is correct.
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