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Thread: Non-Uniform Circular Motion

  1. #1
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    Non-Uniform Circular Motion

    Hi. I'm having trouble with this circular motion question. I'll post the full question and then my queries beneath it.

    In a “pinewood derby,” children build wooden cars to race on a curved track as shown above. The cars all start at an initial height of 1m. Pulled by gravity (g=9.8 m/s2 ), they travel down a curved track onto a long flat straightaway. The first car across the finish line (at the end of the straightaway) wins. There is a strict limit on the weight of each car: 150 g.

    Let us suppose that the initial ramp is shaped like one fourth of a circle with radius=1m. A car is observed at a height of 0.5 m with a speed of 3 m/s.
    1. What is the acceleration of the car at this moment?
    2. What is the normal acceleration (the normal force / the mass of the car) on the car at this moment?
    3. If the coefficient of friction for the car on the track is µ=0.02, what is the acceleration due to friction at this moment?

    1. I simply did $\displaystyle a=\frac{v^2}{r} = 9 ms^{-1}$, as at this moment we can treat it as uniform circular motion(?); however, does gravity need to be considered here?
    2. I said $\displaystyle N=mg{\cdot}cos(45)+\frac{mv^2}{r}$ and went from there, but I'm unsure if this is the right approach.
    3. I know that $\displaystyle F_{r}={\mu}N$, and that it must be in the opposite direction to the net force which is not(?) towards the centre of the circle at this point, but instead in the direction of motion. As such I did $\displaystyle ma={\mu}N$ and solved for a , but I'm really unsure about this.

    Any help would be much appreciated.
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  2. #2
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    Re: Non-Uniform Circular Motion

    for no. 1, the value of 9 that you calculated is the centripetal acceleration. however there is a transverse acceleration, acting in the direction perpendicular to the centripetal acceleration. the vector sum of these 2 gives the overall acceleration of the whole body. if you have not learned about transverse acceleration, then i guess you are expected to give the centripetal acceleration as you did.

    no. 2, i agree with using $\displaystyle mgcos\theta+\frac{mv^2}{r}$. but how did you know that the angle is 45? i could be wrong, but the angle i obtained is 60 degree from the vertical. check your trigonometry from the sketch of the circle. and N that u calculated is the force. remember to divide by m to get the acceleration. specify the direction of this acceleration.

    no. 3, i think your method is correct but perhaps you don't clearly understand $\displaystyle F=\mu R$. in $\displaystyle F=\mu R$, F is the limiting frictional force acting when normal force(perpendicular to the friction) $\displaystyle R$ acts . in this case, your $\displaystyle N$ that you have calculated in 2, is the normal force on the car by the track, which is perpedicular to the track. and your friction acts along the track. so it's clear that $\displaystyle F_{fric}=\mu N$. like wise this is force; divide by m to get acceleration. specify the direction of this acceleration.
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