3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
1 answer:
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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