A ball is swung from a 5 meter rope that is attached at a central point.
The ball's velocity is 10m/s. What is the centripetal acceleration of the
ball?
2006-10-08 18:02:44 · 4 answers · asked by .......... 4 in Science & Mathematics ➔ Physics
20m/s^2
The formula is v^2/r
10^2/5=20
2006-10-08 18:12:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
A ball moves along a circular path with a 5 m radius (r) at a velocity of 10 m/s (v). How much acceleration (a_n) is required normal to the balls velocity vector to keep it moving along the circle?
Since a_n = v²/r (from any elementary physics text) the acceleration is 20 m/s²
Doug
2006-10-09 01:13:09 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
Isn't the formula for centripetal acceleration for circular motion: a = v^2/r
I think your answer is: 10^2/5 = 20 m/s/s
2006-10-09 01:08:57 · answer #3 · answered by tbolling2 4 · 1⤊ 0⤋
a = w^2*r, where w = angular velocity, The angular velocity is tangential velocity vt / r, so a = (vt^2/r^2)*r, or vt^2/r, here r = 5m, vt=10m/s
2006-10-09 01:09:47 · answer #4 · answered by gp4rts 7 · 0⤊ 4⤋