Curvilinear Problem Example 2

A car starts from rest at point A and increases its speed at a constant rate as it enters a curve. The magnitude of the total acceleration of the car as it passes point B, which is $\frac{400}{3} m$ along the curve from point A, is $2.5 \frac{m}{s^2}$. The radius of curvature at point B is 200 m.

a.) Determine the normal and tangential components of the acceleration at point B Answer»

b.) If the tangential acceleration at point B is directed downward to the right at an angle of $36.87^\circ$ below the horizontal and the normal acceleration is also directed downward, determine the rectangular components of the acceleration at point B. Answer»

SOLUTION:

Step 1: Draw

Just like in any of our problems, we try to visualize first the problem before solving it.

Step 2: Write the Problem

1. normal acceleration, $a_{N}$

2. tangential acceleration, $a_{tan}$

Step 3: Guess

The first thing you wanna ask yourself is: what are some equations do I know given the problem?

For me, the first equation I can think of is: $$a_{total}^{2} = a_{tan}^{2} + a_{N}^{2}$$

at point B,

After writing equation above, I’m stuck. But, when I think about it, I haven’t utilized all the given values:

Recall if I could just imagine bending the curved path into a straight line[put image], then, perhaps, I could the constant acceleration equations I’ve previously known. $$v = v_{o} + at$$ $$v^2 = v_{o}^2 + 2a\Delta s$$ $$s = s_{0} + v_ot + \frac{1}{2}at^2$$

Caveat: the acceleration to be used should be the acceleration along the curve.Hence, tangential acceleration more info

As the time is not given, the most logical first equation is: $$v^2 = v_{o}^2 + 2a_{tan}\Delta s$$

We set the point A as the initial point and point B as the final point, hence:

We have one more given that we haven’t utilized, radius of curvature at point B. It might be: $$a_{N} = \frac{v^2}{\rho}$$

Substitute equation (3) to (2), we get $$200 a_{N} = 2a_{tan} \left( \frac{400}{3} \right)$$

Substituting equation (4) to (1), $$2.5^{2} = a_{tan}^{2} + \left(\frac{4}{3} a_{tan} \right)^2$$

from here we get, $$a_{tan} = 1.5 \frac{m}{s^2}$$

Using equation (1) and solving for $a_{N}$, $$2.5^{2} = 1.5^2 + a_{N}^{2}$$ $$a_{N} = 2.0 \frac{m}{s^2}$$

Step 4: Check if all problems are met

No, as we still have to solve b.

Step 4.1: Visualize

Step 4.2: write the problem

$\vec{a_{x}}, \vec{a_{y}}$

Step 4.3: Guess

Just by looking at the diagram above, it seems that what we need to do is to break down the tangential component $a_{tan}$ into its x- and y-component. Similarly, do the same procedure with the normal component

By inspection, $$\vec{a_{tan}} = |\vec{a_{tan}}|\cos(-36.87^\circ) \hat{\imath} + |\vec{a_{tan}}|\sin(-36.87^\circ) \hat{\jmath}$$

Note: $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = \sin(\theta) $, $$\vec{a_{tan}} = |\vec{a_{tan}}|\cos(36.87^\circ) \hat{\imath} - |\vec{a_{tan}}|\sin(36.87^\circ) \hat{\jmath}$$

For the normal component $\vec{a_{N}}$,
By visual inspection, the angle between x- and y- component is $ 90^\circ$, hence the angle between the tangential component and y-component is equal to $ 90^\circ - 36.87^\circ = 53.13^\circ$

It should be noted that the angle between tangential and normal component is equal to $ 90^\circ$, this means that the angle between the normal component and y-component is equal to $ 90^\circ - 53.13^\circ = 36.87^\circ$

Therefore, $$\vec{a_{N}} = -|\vec{a_{N}}|\sin(36.87^\circ) \hat{\imath} - |\vec{a_{N}}|\cos(36.87^\circ) \hat{\jmath}$$

So, to get the $a_{x}$ we simply get the sum of the x-components of equation (5) and (6),

Similarly,