- #1
unknown_2
- 28
- 0
This problem has been discussed before, but one of the parts were not discussed.
Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t).
r(t) = Rcos(\omega t)\textbf{i} + Rsin(\omega t)\textbf{j}
v = \omega R
a = \frac{v^2}{R}
a = \omega v
v = \frac{d}{dt}r(t) \times \omega
btw, this is a Mastering engineering question, so i submitted and it says that:
The correct answer does not depend on the variables and functions: d, d, r.
is there another way to display this or am i missing something?
thx,
Homework Statement
Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t).
r(t) = Rcos(\omega t)\textbf{i} + Rsin(\omega t)\textbf{j}
Homework Equations
v = \omega R
a = \frac{v^2}{R}
a = \omega v
The Attempt at a Solution
v = \frac{d}{dt}r(t) \times \omega
btw, this is a Mastering engineering question, so i submitted and it says that:
The correct answer does not depend on the variables and functions: d, d, r.
is there another way to display this or am i missing something?
thx,
