Question

A wheel rotates around a stationary axis so that the rotation
angle p varies with time as cp = ate, where a = 0.20 rad/s2. Find the
total acceleration w of the point A at the rim at the moment t = 2.5 s
if the linear velocity of the point A at this moment v = 0.65 m/s.
1.45. A shell acquires the initial velocity v = 320 m/s, having
made n = 2.0 turns inside the barrel whose length is equal to 1 =
= 2.0 m. Assuming that the shell moves
inside the barrel with a uniform accelera- A
tion, find the angular velocity of its axial
rotation at the moment when the shell
escapes the barrel.

Answer 1

Answer:

Here, angle is, φ=at2

So, angular speed is, Ω=dtdφ=2at=2×0.2t=0.4t

and, angular accelration is, α=dtdΩ=2a=0.4  rad/s2

So linear velocity is, v=ΩR⟹0.4tR=0.65

At t=2.5s, 

0.4×2.5×R=0.65⟹R=0.65 m

So tangential acceleration is at=αR=0.4×0.65=0.26 m/s2

Radial acceleration is ar=v2/R=(0.65)2/0.65=0.65 m/s2

Net acceleration is ω=0.262+0.652=0.7

So, 10ω=7

I hope this will help you

have a nice day

Answer 2

Explanation:

here is your answer...

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