What Is the Final Velocity of a Car After Descending a Semicircular Track?

[drizzt56]

Homework Statement

Real life setup:
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I have a car whose mass is 500g (.5kg) at the top of the left plane rolling down this stage - what is its velocity after it passes the curve at the bottom (that is, if it can even pass the curve)?

That left plane is 130m, elevated 45 degrees. That curve at the bottom is suppose to be half of a circle with a diameter of 30m. The right and flat planes are 30m, but that's irrelevant for now.

Homework Equations

The Attempt at a Solution

The force down the 130cm plane = mg * sin45 = 3.54N (mg = .5kg*10)
The coefficient of rolling friction I got from some sources for my car accumulates to about .12, so 3.54N - .12 * (.5kg *10 * cos45) = .4
3.54N - .4N = 3.5N down the plane
F = ma, 3.54N = .5kg(a)
a = 7.1m/s^2
final velocity = sqrt(initial velocity^2 + 2*a*d), = sqrt(0 + 2*7.1*1.3m) = 4.3m/s

So, after it rolls down that left plane, just before approaching onto the curve, its final velocity is 4.3m/s
I have no idea what to do after this... At first, I thought it'd be something with centripetal force, but then I realized the car's velocity won't be constant (or at least I don't think so). Any input on what to do afterward will help =/

 

[rl.bhat]

1.

So, after it rolls down that left plane, just before approaching onto the curve, its final velocity is 4.3m/s
I have no idea what to do after this... At first, I thought it'd be something with centripetal force, but then I realized the car's velocity won't be constant (or at least I don't think so). Any input on what to do afterward will help =/

When the car enters the curve, it has a velocity vi. If there is no friction, you can write

PE_i + KE_i = PE_f + KE_f.

Since the friction is present, the equation becomes

PE_i + KE_i = PE_f + KE_f + the work done by the frictional force...(1)

The frictional force depends on the normal reaction. And here normal reaction is not constant.

The contribution to the normal reaction is 1) component of the weight and 2) the reaction to the centripetal force.

So the frictional force

$$f_r = \mu(mg\cos{\theta} + \frac{mv^2}{r})$$ and

net downward force = mgsin(θ) - f_r

For a small displacement ds at any point on the curve, the work done is given by

But ds = r*d$$\theta$$

$$W = (mg\sin{\theta} - f_r)ds = (mg\sin{\theta})rd{\theta} - \mu(mg\cos{\theta} + \frac{mv^2}{r})rd{\theta}$$

Substitute these in the equation (1) and find the integration from θ = π/2 to 0 degree

Find Ε_f and from that find the final velocity at the bottom of the curve.