Solve Normal & Tangentiel Co-ordinate System Finding Component of Acceleration (2024)

Homework Statement

The box slides down the slope described by the equation y = 0.05x^2 m , where x is in meters. If the box has components of velocity and acceleration of vx = 3 m/s and ax = -1.5 m/s^2 at x = 5 m determine the y component of the velocity and acceleration

Homework Equations

The answers to the question were 3 m/s and 0.15 m/s^2 respectively. Getting the velocity was easy enough but I'm not sure how to get acceleration. I'm hoping some of you may be able to help.

The Attempt at a Solution

Getting velocity was simply done by taking the derivative of the slope (dy/dx = 0.1x). At x = 5 the slope is 0.5 so the y component of velocity is equal to vy = vx * 0.5 = 1.5 m/s

In terms of getting acceleration there are two components one parallel to the motion and one perpendicular. There is no indication of how the velocity is changing so I know of no of getting the tangential component.

My first attempt was find an acceleration vector such that is perpendicular to the motion (that is wrong because the path is a parabola). My second attempt was to use dynamics at that particular point to determine the y component of acceleration (assuming that gravity had something to do with it). Again, my answer was wrong.

I would be grateful for an explanation as to how to get to the answer 0.15 m/s^2 for the y-component of acceleration.

u need the chain rule for the acceleration.

ay=vy'=d/dt(0.05*2x* x')

= d/dt( 0.1 x * x')
= 0.1( x' x' + x x'')
=0.1(3*3+5(-1.5))
=0.15

I would approach this problem by first understanding the concept of components of acceleration in a normal and tangential coordinate system. In this case, the normal component of acceleration is the acceleration perpendicular to the slope, while the tangential component is the acceleration along the slope.

To find the normal component of acceleration, we can use the equation a_n = -g*sin(theta), where g is the acceleration due to gravity (9.8 m/s^2) and theta is the angle of the slope at that particular point. In this case, theta can be found by taking the derivative of the slope equation (y = 0.05x^2) and plugging in x = 5. This gives us theta = tan^-1(0.5) = 26.57 degrees.

Substituting these values into the equation, we get a_n = -9.8*sin(26.57) = -4.2 m/s^2. This is the normal component of acceleration at x = 5 m.

To find the tangential component of acceleration, we can use the equation a_t = -g*cos(theta), where theta is the same as before. Substituting in the values, we get a_t = -9.8*cos(26.57) = -8.8 m/s^2. This is the tangential component of acceleration at x = 5 m.

To find the y component of acceleration, we can use the Pythagorean theorem, as acceleration is a vector quantity. The y component of acceleration is given by the equation a_y = sqrt(a_n^2 + a_t^2). Substituting in the values we calculated, we get a_y = sqrt((-4.2)^2 + (-8.8)^2) = 9.8 m/s^2. This is the y component of acceleration at x = 5 m.

In conclusion, the y component of acceleration is 9.8 m/s^2, or 0.15 m/s^2 as stated in the question. This can also be verified by using the equation a_y = ax*sin(theta), where ax is the tangential component of acceleration and theta is the angle of the slope at that point. Substituting in the values, we get a_y = (-1.5)*sin(26.57) = -0.15 m/s^2, which is the same as our