If I ride a bike through a loop-the-loop with R=3m, what minimum speed must I

J

JOE A

New member
have at the top of the loop? well that's it really, I can't grasp how to do this problem and would like some assistance please.
 
Hi Joe,

Here's your answer.

v = velocity = ???

R = radius of loop = 3 m

g = gravitational acceleration = 9.81 m/s^2


At the top of the loop, in order for you to stay on the track the centripetal force, F(cent), must complete balance your weight (W):

F(cent) = W


We recall that your weight is calculated by: W = m*g (m = mass of you + bike)

And that centripetal force is: F(cent) = m*v^2 / R

So:

m*g = m*v^2 / R

Solving for v:

v = sqrt(g*R)

= sqrt[(9.81m/s^2)*(3m)]

= 5.42 m/s


So you would have to travel at 5.42 m/s in order to stay on the track.


That's the answer!






aNAgah.
 
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