Circular momentum - Loop-the-loop question.

[Ripe]

Homework Statement

A loop-the-loop machine has radius 18m. What is the minimum speed at which a cart must travel so that it will safely loop the loop?

Homework Equations

centripetal acceleration: v^2/r
velocity=(2r*pi)/t
Perhaps more?

The Attempt at a Solution

To be honest, my main problem is that I'm not really sure where to start. I calculated the displacement a cart must travel, i.e. the circumference of the diameter, but I don't know what to do next..

 

[LawrenceC]

You want gravity to be nullified by centripetal acceleration when car is at highest point.

 

[Ripe]

Ah. You mean that when it is at the highest point, the centriputal acceleration will have the same magnitude and direction as gravity and thus, we can deduce that the centriputal acceleration is 9.8ms^-2?

 

[LawrenceC]

That's it.

 

[Nickie]

I would approach this problem by first identifying the key equations and concepts that are relevant to the question. In this case, we can use the equations for centripetal acceleration and velocity to determine the minimum speed needed for the cart to safely complete the loop-the-loop.

To start, we can use the formula for centripetal acceleration, a = v^2/r, where a is the centripetal acceleration, v is the velocity, and r is the radius of the loop-the-loop. We can rearrange this equation to solve for the velocity, v, which gives us v = √(ar).

Next, we can use the formula for velocity in circular motion, v = 2πr/t, where t is the time it takes for the cart to complete one full revolution around the loop-the-loop. We can rearrange this equation to solve for the time, t, which gives us t = (2πr)/v.

Now, we can substitute the expression for v from the first equation into the second equation to get t = (2πr)/√(ar). This gives us the time it takes for the cart to complete one full revolution around the loop-the-loop, in terms of the radius and acceleration.

Finally, we can use the given radius of 18m and the acceleration due to gravity, g = 9.8 m/s^2, to solve for the minimum velocity needed for the cart to safely complete the loop-the-loop. Plugging in these values into the equation for v, we get v = √(9.8*18) = 13.4 m/s.

Therefore, the minimum speed at which the cart must travel in order to safely complete the loop-the-loop is 13.4 m/s. This speed is necessary to provide enough centripetal acceleration to keep the cart moving in a circular path without falling off the track.