What is the potential energy of a 3 kilogram-ball that is on the ground?
2 answers:
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The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
Know recall that:
We now need to know how much distance is traveled during a minute or 60 seconds:
Finally we divide this result with t=60 seconds:
Where the distance units were named units as the length unit is not specified in this exercise.<span />
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
Know recall that:
We now need to know how much distance is traveled during a minute or 60 seconds:
Finally we divide this result with t=60 seconds:
Where the distance units were named units as the length unit is not specified in this exercise.<span />
(a)
The relationship beween centripetal acceleration and angular speed is
where
is the angular speed
r is the radius of the circular path
Here we gave
is the centripetal acceleration
r = 5.15 m is the radius
Solving for , we find:
(b) 21.3 m/s
The relationship between the linear speed and the angular speed is
where
v is the linear speed
is the angular speed
r is the radius of the circular path
In this problem we have
r = 5.15 m
Solving the equation for v, we find