Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer,
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.
1) 1.2 m/s
First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by
where
is the initial angular velocity
is the angular acceleration
Substituting t = 0.200 s, we find
Let's now convert it into rad/s:
The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:
And so now we can find the tangential speed at t = 0.200 s:
2)
The tangential acceleration of a point rotating at a distance r from the centre of the circle is
where is the angular acceleration.
First of all, we need to convert the angular acceleration into rad/s^2:
A point on the tip of the blade has a distance of
r = 0.400 m
From the centre; so, the tangential acceleration is
3)
The centripetal acceleration is given by
where
v is the tangential speed
r is the distance from the centre of the circle
We already calculate the tangential speed at point a):
v = 1.2 m/s
while the distance of a point at the end of the blade from the centre is
r = 0.400 m
Therefore, the centripetal acceleration is