A rotating rod that is 15.3 cm long is spun with its axis through one end of the rod so that the other end of the rod has a speed of 2000 m/s (4,500 mph).
a. What is the centripetal acceleration of the other end of the rod?
b. If you were to attach a 1.0-g object to the end of the rod, what force would be needed to hold it on the rod?
Please help me with this problem.
I'm having a hard time understanding it.
2006-11-27 14:36:53 · 1 answers · asked by swimmertommy 1 in Science & Mathematics ➔ Physics
Centripetal acceleration A = w^2*r, where v is velocity, w is the rotation rate in rad/s and r is the radius. Since velocity v = w*r, we can substitute v/r for w in the equation for A: A = v^2/r.
So in this case, A = 2000^2/0.153 = 2.6144 * 10^7 m/s^2 (the answer to (a). For (b), F = M*A = .001 * A = 26144 N.
2006-12-01 01:27:09 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋