Respuesta :
a) 219.8 rad/s
b) [tex]20.0 rad/s^2[/tex]
c) [tex]2.9 m/s^2[/tex]
d) [tex]7005 m/s^2[/tex]
e) Towards the axis of rotation
f) [tex]0 m/s^2[/tex]
g) 31.9 m/s
Explanation:
a)
The angular velocity of an object in rotation is the rate of change of its angular position, so
[tex]\omega=\frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement
t is the time elapsed
In this problem, we are told that the maximum angular velocity is
[tex]\omega_{max}=35 rev/s[/tex]
The angle covered during 1 revolution is
[tex]\theta=2\pi[/tex] rad
Therefore, the maximum angular velocity is:
[tex]\omega_{max}=35 \cdot 2\pi = 219.8 rad/s[/tex]
b)
The angular acceleration of an object in rotation is the rate of change of the angular velocity:
[tex]\alpha = \frac{\Delta \omega}{t}[/tex]
where
[tex]\Delta \omega[/tex] is the change in angular velocity
t is the time elapsed
Here we have:
[tex]\omega_0 = 0[/tex] is the initial angular velocity
[tex]\omega_{max}=219.8 rad/s[/tex] is the final angular velocity
t = 11 s is the time elapsed
Therefore, the angular acceleration is:
[tex]\alpha = \frac{219.8-0}{11}=20.0 rad/s^2[/tex]
c)
For an object in rotation, the acceleration has two components:
- A radial acceleration, called centripetal acceleration, towards the centre of the circle
- A tangential acceleration, tangential to the circle
The tangential acceleration is given by
[tex]a_t = \alpha r[/tex]
where
[tex]\alpha[/tex] is the angular acceleration
r is the radius of the circle
Here we have
[tex]\alpha =20.0 rad/s^2[/tex]
d = 29 cm is the diameter, so the radius is
r = d/2 = 14.5 cm = 0.145 m
So the tangential acceleration is
[tex]a_t=(20.0)(0.145)=2.9 m/s^2[/tex]
d)
The magnitude of the radial (centripetal) acceleration is given by
[tex]a_c = \omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular velocity
r is the radius of the circle
Here we have:
[tex]\omega_{max}=219.8 rad/s[/tex] is the angular velocity when the fan is at full speed
r = 0.145 m is the distance of the gum from the centre of the circle
Therefore, the radial acceleration is
[tex]a_c=(219.8)^2(0.145)=7005 m/s^2[/tex]
e)
The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.
Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.
f)
The magnitude of the tangential acceleration of the fan at any moment is given by
The tangential acceleration is given by
[tex]a_t = \alpha r[/tex]
where
[tex]\alpha[/tex] is the angular acceleration
r is the radius of the circle
When the fan is rotating at full speed, we have:
[tex]\alpha=0[/tex], since the fan is no longer accelerating, because the angular velocity is no longer changing
r = 0.145 m
Therefore, the tangential acceleration when the fan is at full speed is
[tex]a_t=(0)(0.145)=0 m/s^2[/tex]
g)
The linear speed of an object in rotational motion is related to the angular velocity by the formula:
[tex]v=\omega r[/tex]
where
v is the linear speed
[tex]\omega[/tex] is the angular velocity
r is the radius
When the fan is rotating at maximum angular velocity, we have:
[tex]\omega=219.8 rad/s[/tex]
r = 0.145 m
Therefore, the linear speed of the gum as it is un-stucked from the fan will be:
[tex]v=(219.8)(0.145)=31.9 m/s[/tex]
