Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnit
1 answer:
Circular acceleration has 2 components, the centripetal and tangential acceleration.
The formula for calculating the centripetal acceleration of a particle is:
= v^2/r
Where v is the velocity or instantaneous speed and r is the radius.
= (3^2)/6 = 1.5 m/s^2
Assuming that the particle stays on the same circular path, another thing it can have is the tangential acceleration. While the centripetal acceleration is constant, tangential acceleration depends on the location of the particle in the circular path. So to answer A and B, yes we can have an acceleration of 6 and 4 respectively.
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Answer:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:
If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.