At a given instant, the total acceleration of a particle moving clockwise in a circle of radius 2.05 m. The magnitude of the total acceleration of the particle at the given instant is 14.8 m/s^2; the angle between the vector of total acceleration and the radius-vector of the particle is 30 degrees. At this instant of time, find the centripetal acceleration of the particle. Answer in units of m/s2.
2.Find the speed of the particle at the samemoment. Answer in units of m/s.
3.Find particle's tangential acceleration at the same moment. Answer in units of m/s^2.
Can someone do this problem and tell me what u get for the answer. I have all the formulas but for some reason i keep getting the wrong answer.
2006-10-20 17:23:43 · 2 answers · asked by polo_05_polo 1 in Education & Reference ➔ Homework Help
A particle cannot travel in a circle if it has a tangential component of acceleration.
1.
The centripetal acceleration will be 14.8cos30°
2.
Insufficient information, or too much information. Assuming the tangential component of acceleration was just applied, and the centripetal acceleration previously was constant,
v^2/r = 14.8cos30°, and
v = sqrt(2.05*14.8cos30°)
3.
The tangential acceleration will be 14.8sin30°
2006-10-20 17:56:37 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
I do not have the patience to type it out but I can try helping on IM using voice.IM me now I am free.
2006-10-20 18:06:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋