Physics Help Please?
There is a merry go round that rotates at a rate of 6.20 revolutions per minute and you are riding on an animal 5.0 m from the axis of rotation. The angular velocity is 0.65 rad/s, the tangential speed is 3.25 m/s, and the centripetal acceleration is 2.11 m/s^2. The carousel slows down so the tangential acceleration is now -0.250 m/s^2.
1. What is the tangential velocity in m/s 3.0 seconds after the carousel begins to slow down?
2. What is the centripetal acceleration in m/s^2 3.0 seconds after the carousel begins to slow down?
3. What is the magnitude of your total acceleration 3.0 seconds after the carousel begins to slow down?
Thank you so much! I worked out all the numbers to the first part, but now I'm stuck when the carousel starts slowing down.
- NCSLv 78 months agoFavorite Answer
1. You are given the initial tangential velocity v, the deceleration a, and the time.
final velocity V = v + a*t
2. Use V from part 1: centripetal a_c = V² / R
3. Total acceleration = √(a² + a_c²)
where a was given as -0.250 m/s².
Hope this helps!
- oldschoolLv 78 months ago
We know that V = ωr = 0.65rad/s*5m = 3.25m/s
mV²/r = the outward force so the outward acceleration
F/m = a = V²/r = 3.25²/5 = 2.1125m/s²
a = αr = -0.25 so α = -0.05 = -1/20 rad/s²
ω = ωi + αt = 0.65 -1/20 *3 = 0.5rad/s = V/r = V/5
V = 2.5rad/s <<<<< 1)
V²/r = 25/4 * 1/5 = 25/20 = 5/4 = 1.25m/s² <<<<< 2)
Total acceleration at 3s = √(0.25²+1.25²) = 1.27 <<<< 3)
- Anonymous8 months ago
Convert tangential velocity to acceleration, square result. Simple.
- 8 months ago
Do your own research and homework. Thank you so much! You will not learn if we answer.